Question 1

How many 7 digit numbers can be formed using all the digits 1, 2, 0, 2, 4, 2 and 4 exactly as given, so that no extra repetition is allowed and leading zero is not permitted?

Accepted Answer

Introduction / Context: This question tests your understanding of permutations of a multiset (repeated digits) with an additional restriction that a valid 7 digit number cannot begin with zero. We must count all distinct permutations of the given digits and then subtract those arrangements that start with zero. This type of problem is very common in aptitude tests involving digit arrangements. Given Data / Assumptions: Available digits: 1, 2, 0, 2, 4, 2, 4. Total digits = 7. Digit 2 appears three times. Digit 4 appears twice. Digits 0 and 1 appear once each. We must form valid 7 digit numbers, so the leading digit cannot be 0. Concept / Approach: First, we count the total number of distinct permutations of all 7 digits treating repeated digits properly. This uses the multiset permutation formula 7! divided by factorials of the counts of repeated digits. Then, we separately count how many of these permutations start with 0 by fixing zero in the first position and permuting the remaining 6 digits. Subtracting the bad arrangements (with leading zero) from the total gives the count of valid 7 digit numbers. Step-by-Step Solution: Step 1: Count total permutations of the 7 digits without any restriction. Step 2: Total permutations = 7! / (3! * 2!) because digit 2 repeats three times and digit 4 repeats twice. Step 3: 7! = 5040, 3! = 6 and 2! = 2, so denominator = 6 * 2 = 12. Step 4: Total permutations = 5040 / 12 = 420. Step 5: Now count permutations that start with 0. Fix 0 in the first position. Step 6: Remaining digits are 1, 2, 2, 2, 4, 4 (six digits with 2 repeated three times and 4 repeated twice). Step 7: Permutations of remaining digits = 6! / (3! * 2!). Step 8: 6! = 720 and 3! * 2! = 6 * 2 = 12, so we get 720 / 12 = 60. Step 9: These 60 arrangements are invalid 7 digit numbers because they begin with 0. Step 10: Valid 7 digit numbers = total permutations - invalid leading zero permutations = 420 - 60 = 360. Verification / Alternative check: As a quick check, note that the total count 420 includes all possible ways to order the 7 digits, regardless of the first digit. Since 1, 2 and 4 are non zero, there should be a substantial number of valid numbers. Excluding 60 sequences that start with 0 leaves 360, which is less than the total but still a large fraction, which is reasonable. A more detailed check could group valid numbers by the first digit (1, 2 or 4) and verify symmetry ideas, but the computed arithmetic from factorial formulas is already very reliable. Why Other Options Are Wrong: The value 120 is far too small and might come from misusing factorials or ignoring repeated digits. The value 240 could arise from subtracting a wrong number of invalid arrangements. The value 424 is larger than the maximum possible distinct permutations of 420, so it cannot be correct. Only 360 correctly equals 420 minus 60 and respects both the multiset permutations and the leading digit restriction. Common Pitfalls: Common mistakes include ignoring that digit 2 and digit 4 are repeated, leading to using 7! instead of 7! / (3! * 2!). Another frequent error is forgetting to remove arrangements that start with 0, which would allow invalid 7 digit numbers with a leading zero. Some students also incorrectly divide by 2 a second time or mix up the counts of repeated digits when calculating 6! / (3! * 2!). Carefully identifying repeated digits and applying constraints step by step prevents these issues. Final Answer: The number of distinct 7 digit numbers that can be formed is 360.

Question 2

The Indian cricket team has 16 players including 2 wicket keepers and 5 bowlers. In how many ways can a cricket eleven be selected if the team must contain exactly 1 wicket keeper and at least 4 bowlers?

Accepted Answer

Introduction / Context: This question tests multiple combination ideas together. The team selection has role based constraints: exactly one of the two wicket keepers must be chosen, and the number of bowlers must be at least 4 out of 5 available bowlers. The remaining players come from the other group of non bowler and non wicket keeper players. We tackle the problem by splitting into cases based on how many bowlers are chosen and then summing the counts. Given Data / Assumptions: Total players = 16. Wicket keepers = 2. Bowlers = 5. Other players (neither wicket keeper nor bowler) = 16 - 2 - 5 = 9. The team size is 11 players. Exactly 1 wicket keeper must be selected. At least 4 bowlers must be selected. Concept / Approach: We first choose 1 wicket keeper from 2. Then we consider two possible cases for the bowlers: choosing exactly 4 bowlers from 5, or choosing all 5 bowlers. For each case, the remaining players needed to complete 11 are chosen from the 9 other players. We use combinations for each selection and multiply where appropriate. Finally, we sum the counts from the two cases to get the total number of valid teams. Step-by-Step Solution: Step 1: Choose the wicket keeper. Ways = 2C1 = 2. Step 2: Case 1: Team has exactly 4 bowlers. Step 3: Choose 4 bowlers from 5: 5C4 = 5. Step 4: So far selected players = 1 wicket keeper + 4 bowlers = 5. Step 5: Remaining players needed = 11 - 5 = 6, chosen from the 9 other players. Step 6: Ways to choose these 6 players = 9C6 = 84. Step 7: Total teams in Case 1 = 2 * 5 * 84 = 2 * 420 = 840. Step 8: Case 2: Team has all 5 bowlers. Step 9: Choose 5 bowlers from 5: 5C5 = 1. Step 10: So far selected players = 1 wicket keeper + 5 bowlers = 6. Step 11: Remaining players needed = 11 - 6 = 5, chosen from the 9 other players. Step 12: Ways to choose these 5 players = 9C5 = 126. Step 13: Total teams in Case 2 = 2 * 1 * 126 = 252. Step 14: Total valid teams = Case 1 + Case 2 = 840 + 252 = 1092. Verification / Alternative check: You can verify the combination values individually: 9C6 = 84 and 9C5 = 126 can be checked using factorial expressions or known combinatorial symmetry 9C6 = 9C3 and 9C5 = 9C4. Also, the sum of bowlers in all teams should logically range only over 4 or 5, which we have fully covered. There is no overlap between Case 1 and Case 2 because the number of bowlers is different. Therefore, adding the counts is valid, and 1092 is consistent. Why Other Options Are Wrong: Values like 1024, 1900 or 2000 arise from incorrect combinations or forgetting to split into cases for the number of bowlers. For example, counting teams with no restriction on bowlers or using 16C11 directly would ignore the specific constraints and give a completely different value. Only 1092 follows logically from selecting the wicket keeper, then the bowlers, then the remaining players under the given roles. Common Pitfalls: A common mistake is to ignore the requirement of exactly one wicket keeper and at least four bowlers, leading to overcounting or undercounting. Some students treat at least 4 bowlers as exactly 4 bowlers and forget Case 2 where all 5 bowlers are in the team. Another mistake is to choose players in one big step without using separate combinations for each role, which often causes miscounting. Breaking the problem into clear cases and applying the combination formula carefully avoids these errors. Final Answer: The cricket eleven can be selected in 1092 different ways under the given conditions.

Question 3

In how many different ways can 5 different toys be packed into 3 identical boxes such that no box is left empty and any box may hold any number of toys?

Accepted Answer

Introduction / Context: This question involves distributing distinct objects (toys) into identical boxes with a non emptiness condition. When boxes are identical, we cannot distinguish between arrangements that differ only by swapping the boxes. Counting such configurations is related to set partitions and Stirling numbers of the second kind. For 5 distinct toys and 3 identical non empty boxes, the answer equals the Stirling number S(5, 3). Given Data / Assumptions: Number of toys = 5. Each toy is different. Number of boxes = 3. Boxes are identical (not labeled), so their order does not matter. No box may remain empty. Any box may hold one or more toys, as long as the total toys remain 5. Concept / Approach: We are essentially partitioning a set of 5 distinct elements into exactly 3 non empty unlabeled groups. The number of such partitions is the Stirling number of the second kind S(5, 3). This Stirling number can be computed using a recurrence relation, a combinatorial reasoning approach, or a direct formula. For small values, it is also well known and tabulated: S(5, 3) equals 25. Because the boxes are identical, we do not distinguish permutations of the groups themselves. Step-by-Step Solution: Step 1: Recognize that distributing toys into 3 identical boxes with no empty box corresponds to partitions of the set of 5 toys into 3 non empty unlabeled subsets. Step 2: Use the Stirling number of the second kind S(n, k) to count partitions of an n element set into k non empty unlabeled subsets. Step 3: Here, n = 5 and k = 3, so we need S(5, 3). Step 4: Using the recurrence relation S(n, k) = S(n - 1, k - 1) + k * S(n - 1, k), along with known base values, we can compute S(5, 3). Step 5: Known values include S(3, 1) = 1, S(3, 2) = 3, S(3, 3) = 1, S(4, 1) = 1, S(4, 2) = 7, S(4, 3) = 6 and S(4, 4) = 1. Step 6: Compute S(5, 3) = S(4, 2) + 3 * S(4, 3) = 7 + 3 * 6 = 7 + 18 = 25. Step 7: Hence the number of distinct packing patterns is 25. Verification / Alternative check: You can also reason combinatorially about possible distributions of sizes. Since boxes are non empty and identical, the group sizes of toys can only be 3-1-1 or 2-2-1. One may count how many ways to choose which toys go into each group size arrangement, then divide out the permutations of groups of the same size to respect identical boxes. Summing over 3-1-1 and 2-2-1 distributions also leads to 25. This more detailed method is more tedious but matches the Stirling number result, confirming the answer. Why Other Options Are Wrong: The value 36 might arise from treating boxes as distinct or from using a combinations and permutations mix incorrectly. The value 24 often comes from simple factorial thinking like 4!, which does not model the problem structure. The value 72 is far too large for unlabeled boxes and mixes labeled box counting with non emptiness constraints. Only 25 aligns with the correct set partition interpretation using S(5, 3). Common Pitfalls: The most common error is forgetting that the boxes are identical, not labeled. Many students first count distributions into labeled boxes and then forget to divide out the permutations of boxes that represent the same grouping. Another pitfall is allowing empty boxes, which would correspond to partitions into at most 3 subsets instead of exactly 3. Some also confuse this with distributing identical objects, which is a completely different problem. Always check whether objects or boxes are distinct and whether emptiness is allowed before choosing the counting method. Final Answer: The number of different ways to pack the 5 toys into 3 identical non empty boxes is 25.

Question 4

Seven parallel horizontal lines intersect six parallel vertical lines in a plane. How many different parallelograms can be formed using these lines as the sides of the parallelograms?

Accepted Answer

Introduction / Context: This question involves counting parallelograms formed by a grid of parallel lines. Parallelograms are determined by selecting two distinct horizontal lines and two distinct vertical lines. Each pair of horizontal lines and each pair of vertical lines forms a unique parallelogram. Recognizing this pattern and using combinations reduces the geometry question to a simple counting problem. Given Data / Assumptions: Number of horizontal parallel lines = 7. Number of vertical parallel lines = 6. All horizontal lines are parallel to each other and distinct. All vertical lines are parallel to each other and distinct. Horizontal lines are perpendicular to vertical lines. Any two distinct horizontal lines and two distinct vertical lines determine a parallelogram. Concept / Approach: A parallelogram on this grid is defined by choosing two distinct horizontal lines for the top and bottom, and two distinct vertical lines for the left and right sides. Since lines within each family (horizontal or vertical) are parallel, any two horizontal lines combined with any two vertical lines create exactly one unique parallelogram. Therefore, the total number of parallelograms is the product of the number of ways to choose horizontal pairs and the number of ways to choose vertical pairs. Step-by-Step Solution: Step 1: Compute the number of ways to choose 2 horizontal lines from 7. Step 2: Number of horizontal pairs = 7C2 = 7 * 6 / 2 = 21. Step 3: Compute the number of ways to choose 2 vertical lines from 6. Step 4: Number of vertical pairs = 6C2 = 6 * 5 / 2 = 15. Step 5: For each horizontal pair, any vertical pair forms a parallelogram. Step 6: Total parallelograms = 21 * 15. Step 7: Calculate 21 * 15 = 315. Step 8: Therefore, 315 distinct parallelograms can be formed. Verification / Alternative check: A rough check is to see that the number of horizontal pairs 21 and vertical pairs 15 are both positive and moderate. Their product is 315, which is comfortably below the total number of rectangles that could be formed if more directions were allowed. Also, if we had one fewer vertical line (5), the number of parallelograms would be 7C2 * 5C2 = 21 * 10 = 210, and adding another vertical line reasonably increases the count to 315. The arithmetic 21 * 15 = 315 is straightforward and confirms the result. Why Other Options Are Wrong: The value 215 may come from a miscalculation of 7C2 or 6C2 or an incorrect addition instead of multiplication. The value 415 is larger than the correct product and has no natural combinatorial justification here. The value 115 is far too small for a grid with 7 by 6 lines. Only 315 corresponds exactly to 7C2 * 6C2 and matches the logical structure of choosing two horizontal and two vertical lines. Common Pitfalls: Students sometimes confuse the number of rectangles or squares with the number of parallelograms in such problems. In this configuration, every rectangle is also a parallelogram, but the counting method remains the same: pairs of lines in each direction. Another mistake is to try to count parallelograms by visual inspection, which quickly becomes messy. Others may forget that the lines are infinite and not counted by segments. Using combinations for line pairs is the clean and reliable method. Final Answer: The total number of parallelograms formed by the 7 horizontal and 6 vertical lines is 315.

Question 5

In how many distinct ways can 8 beads of different colours be strung to form a necklace if arrangements that can be obtained from one another by rotation or reflection are considered the same?

Accepted Answer

Introduction / Context: This problem is another necklace counting question involving symmetry. We have 8 distinct beads and we are forming a circular necklace, where arrangements that are the same under rotation or reflection are considered identical. This requires using the formula for the number of distinct necklaces with n different beads, taking both rotations and reflections into account. Given Data / Assumptions: Number of beads = 8. All beads are different in colour. Beads are arranged in a circular necklace. Rotations of a bead arrangement are considered the same necklace. Reflections (mirror images) of a bead arrangement are also considered the same necklace. Concept / Approach: For n distinct beads arranged in a circle, the number of circular permutations when only rotations are identified is (n - 1)!. When reflections are also identified, each unique circular arrangement is counted twice in this count (once for each orientation), so we divide by 2. Therefore, the number of distinct necklaces equals (n - 1)! / 2 when both rotations and reflections are treated as equivalent for n distinct beads. Step-by-Step Solution: Step 1: Let n = 8 for the 8 distinct beads. Step 2: First handle rotational symmetry. The number of circular permutations ignoring reflection is (n - 1)! = 7!. Step 3: Compute 7! = 5040. Step 4: Now include reflection symmetry. In a necklace, each arrangement has a mirror image that looks the same when flipped, so every distinct necklace is counted twice in the 7! count. Step 5: Divide by 2 to correct for the double counting: distinct necklaces = 7! / 2. Step 6: Calculate 7! / 2 = 5040 / 2 = 2520. Step 7: Therefore, there are 2520 distinct necklaces. Verification / Alternative check: We can test the general formula by using smaller numbers of beads. For instance, with n = 4 distinct beads, the formula gives (4 - 1)! / 2 = 3! / 2 = 3 distinct necklaces, which can be verified by drawing and enumerating possible patterns up to rotation and reflection. This matches known results. Similarly, for n = 3 beads, the formula gives 1 necklace, which is also intuitive. These checks support using (n - 1)! / 2 for n = 8, giving 2520. Why Other Options Are Wrong: The value 5040 is 7! and accounts only for rotation but not for reflection symmetry. The value 2880 does not match any standard division of 7! and indicates an incorrect adjustment. The value 4320 is also inconsistent with the required division by 2. The value 1260 is exactly half of 2520 and would be correct only if we had already divided by 2 for another symmetry incorrectly. Only 2520 corresponds to 7! / 2 and properly includes both rotational and reflection equivalence. Common Pitfalls: Some students mistakenly use 8! for linear arrangements or 7! for circular arrangements without considering reflections. Others may incorrectly divide by 8 instead of 2, or they apply the reflection correction twice. Another common issue is misunderstanding the phrase arrangements that look the same under rotation and reflection. Always ask whether both types of symmetry are being considered and apply the formula accordingly. For n distinct beads and both symmetries, use (n - 1)! / 2. Final Answer: The number of distinct necklaces that can be formed from 8 different beads is 2520.

Question 6

In how many different ways can a committee consisting of exactly 5 men and 6 women be formed from a group of 8 men and 10 women?

Accepted Answer

Introduction / Context: This is a straightforward committee selection problem, where we must choose a fixed number of men and women from separate pools. Order does not matter, so we rely on combinations rather than permutations. The result comes from choosing men and women independently and then multiplying the combination counts to get the total number of committees. Given Data / Assumptions: Total men available = 8. Total women available = 10. The committee must include exactly 5 men. The committee must include exactly 6 women. Committee membership is unordered; only who is in the committee matters, not any arrangement. Concept / Approach: We treat the selection of men and women as independent tasks. The number of ways to choose 5 men from 8 is 8C5, and the number of ways to choose 6 women from 10 is 10C6. Every selection of men can be combined with every selection of women, so the rule of product tells us to multiply the two combination values. The final answer is 8C5 * 10C6. Step-by-Step Solution: Step 1: Compute 8C5 for choosing 5 men from 8. Step 2: Use the symmetry 8C5 = 8C3. Step 3: 8C3 = 8 * 7 * 6 / (3 * 2 * 1) = 56. Step 4: Compute 10C6 for choosing 6 women from 10, using symmetry 10C6 = 10C4. Step 5: 10C4 = 10 * 9 * 8 * 7 / (4 * 3 * 2 * 1) = 210. Step 6: Multiply the two counts: total committees = 56 * 210. Step 7: First compute 56 * 200 = 11200. Step 8: Then add 56 * 10 = 560, giving 11200 + 560 = 11760. Step 9: Therefore, there are 11760 distinct committees possible. Verification / Alternative check: We can verify the combination values with factorial formulas: 8C5 = 8! / (5! * 3!) and 10C6 = 10! / (6! * 4!). Calculations will confirm 56 and 210 respectively. Also, note that the size of the committee is 11, and it is reasonable that from a pool of 18 people (8 men plus 10 women) there are many thousands of possible committees. The product 56 * 210 fits this expectation and matches both symmetry and factorial based calculations. Why Other Options Are Wrong: The value 53400 is larger and might come from multiplying incorrect combination values. The numbers 17610 and 45000 do not equal 8C5 * 10C6 and reflect arithmetic or formulation errors. The value 60060 is 15C6 and does not correspond to the needed structure. Only 11760 arises from correctly applying the combination formula to both groups and multiplying the results. Common Pitfalls: Some students confuse the problem with permutations and try to use nPr instead of nCr, overcounting arrangements where the same group members are listed in different orders. Others may accidentally choose the wrong number of men or women when setting up the combinations. There is also the possibility of incorrectly treating all 18 people as a single pool and using 18C11 without enforcing gender constraints. Carefully separating the selection of men and women prevents these mistakes. Final Answer: The number of ways to form the committee is 11760.

Question 7

A letter lock consists of three rings, each ring marked with six different letters. What is the maximum number of distinct unsuccessful attempts that can be made to open the lock if the correct combination is unknown?

Accepted Answer

Introduction / Context: This problem is about counting the number of different combinations on a simple letter lock. Each ring can show one of six letters, and the combination consists of the letters shown by the three rings together. We are asked to find the maximum number of incorrect attempts possible, which is simply all possible combinations minus the one correct combination. Given Data / Assumptions: The lock has 3 rings. Each ring has 6 different letters. Every ring can independently show any one of its 6 letters at a time. Exactly one combination is the correct one that opens the lock. All other combinations are unsuccessful attempts. Concept / Approach: The total number of possible combinations is found by multiplying the number of choices for each ring. Since there are 6 choices on each of the 3 rings, the total combinations equals 6 * 6 * 6 which is 6^3. If we assume that each distinct combination is tried at most once, the maximum number of unsuccessful attempts occurs when we try every combination except the correct one. Therefore the number of unsuccessful attempts is total combinations minus 1. Step-by-Step Solution: Step 1: Compute the number of possible letter combinations on the lock. Step 2: For ring 1, there are 6 letter choices. Step 3: For ring 2, there are also 6 letter choices. Step 4: For ring 3, there are again 6 letter choices. Step 5: Total combinations = 6 * 6 * 6 = 6^3 = 216. Step 6: Exactly one of these 216 combinations is the correct one. Step 7: Maximum unsuccessful attempts = total combinations - 1 = 216 - 1 = 215. Verification / Alternative check: An alternative way to view this is to imagine systematically listing all possible three letter sequences using the 6 letters on each ring. There are 6 possibilities for the first letter, 6 for the second, and 6 for the third, giving 216 sequences. If you tried every sequence exactly once, you would be guaranteed to use the correct one at some attempt. The number of wrong ones is simply one less than 216, which is 215, confirming the result. Why Other Options Are Wrong: The value 216 is the total number of possible combinations, including the correct one, so it is too large for the number of unsuccessful attempts. Values like 268 and 254 exceed the total number of combinations or do not have a logical basis here. The value 200 is smaller than the total combinations minus one and implies not all possible incorrect combinations are tried. Only 215 correctly equals 6^3 minus one. Common Pitfalls: Some students forget to subtract 1 and give 216 as the answer, misinterpreting the question as asking for total combinations rather than unsuccessful ones. Others may miscalculate 6^3 or confuse combinations with permutations, although in this case they coincide because the positions (rings) are distinct. Carefully reading that the question is about unsuccessful attempts and recognizing the need for total minus one avoids these issues. Final Answer: The maximum number of distinct unsuccessful attempts to open the lock is 215.

Question 8

An examination has 5 multiple choice questions. The first three questions have 4 answer choices each and the last two questions have 6 answer choices each. How many different answer sequences are possible for the entire paper?

Accepted Answer

Introduction / Context: This question checks basic multiplication principle in counting. For a multiple choice paper, each question can be answered in several ways. The total number of complete answer sequences is the product of the number of choices for each question. Because the first three questions and the last two questions have different numbers of answer choices, we must treat them separately and then multiply their counts. Given Data / Assumptions: Total number of questions = 5. Question 1, 2 and 3 each have 4 answer choices. Question 4 and 5 each have 6 answer choices. The student must choose exactly one answer for each question. All answer choices for different questions are independent. Concept / Approach: We use the fundamental rule of counting, also called the rule of product. If an action can be done in a ways and another independent action can be done in b ways, then the pair can be done in a * b ways. Here, each question is an independent choice. Therefore, the total number of answer sequences equals the product of the number of choices for each question: 4 * 4 * 4 * 6 * 6. Step-by-Step Solution: Step 1: Compute the number of ways to answer the first three questions. Step 2: Each of these three questions has 4 choices, so choices for them = 4 * 4 * 4 = 4^3. Step 3: 4^3 = 64. Step 4: Compute the number of ways to answer the last two questions. Step 5: Each of these two questions has 6 choices, so choices = 6 * 6 = 6^2. Step 6: 6^2 = 36. Step 7: Total answer sequences = ways for first three questions * ways for last two questions. Step 8: Total = 64 * 36. Step 9: Compute 64 * 36 = 64 * (30 + 6) = 64 * 30 + 64 * 6 = 1920 + 384 = 2304. Step 10: Therefore, there are 2304 possible answer sequences. Verification / Alternative check: We can also write the product directly as 4^3 * 6^2 and simplify. Notice that 4^3 = 64 and 6^2 = 36, so 64 * 36 is the same computation as above. Another mental check is to confirm ranges: each of the 5 questions has at least 4 choices and no more than 6, so the total number of answer sequences should lie between 4^5 = 1024 and 6^5 = 7776. The value 2304 is between these bounds, reinforcing that the result is reasonable. Why Other Options Are Wrong: The value 1112 does not equal any simple product of the given powers of 4 and 6 and is too small. The value 1224 is also too small and suggests incomplete multiplication. The value 2426 is not divisible nicely by 4 or 6 and does not arise from any correct product in this context. The value 1024 equals 4^5, which wrongly assumes that all 5 questions have 4 choices. Only 2304 equals 4^3 * 6^2 and is consistent with the problem statement. Common Pitfalls: Students sometimes mistakenly add instead of multiply the number of choices or assume all questions have the same number of options. Another frequent error is to miscalculate 64 * 36, perhaps by incorrectly distributing the product. Always remember that for independent choices, like answering different questions on an exam, we multiply the counts, not add them. Writing the expression systematically as 4^3 * 6^2 and then simplifying helps avoid arithmetic slips. Final Answer: The total number of different answer sequences possible is 2304.

Question 9

In how many distinct ways can all the letters of the word EDUCATION be rearranged to form new words, if every vowel must always occupy a vowel position and every consonant must always occupy a consonant position exactly as in the original word EDUCA…

Accepted Answer

Introduction / Context:This question tests the concept of permutations with positional restrictions. The word EDUCATION contains both vowels and consonants, and we want to rearrange its letters while keeping the pattern of vowel positions and consonant positions the same as in the original word. Such problems are common in aptitude tests to check understanding of permutations and classification of letters into groups like vowels and consonants. Given Data / Assumptions: The word is EDUCATION. The letters are E, D, U, C, A, T, I, O, N. Vowels in this word are E, U, A, I, O. Consonants in this word are D, C, T, N. The relative pattern of vowel and consonant positions must remain exactly the same as in EDUCATION. All letters are used in every arrangement. Concept / Approach:The key idea is that vowels can only occupy positions which were originally vowels in EDUCATION, and consonants can only occupy positions which were originally consonants. Once we fix which positions are for vowels and which are for consonants, we simply count how many ways to permute the vowels among vowel positions and the consonants among consonant positions. Since the choice of vowels is independent of the choice of consonants, the total number of arrangements is the product of the two permutation counts. Step-by-Step Solution:Step 1: Identify positions of vowels in EDUCATION. The word is E D U C A T I O N, so vowels appear at positions 1 (E), 3 (U), 5 (A), 7 (I) and 8 (O). That gives 5 vowel positions.Step 2: Identify positions of consonants. Consonants D, C, T, N appear at positions 2, 4, 6 and 9. That gives 4 consonant positions.Step 3: Count permutations of the 5 distinct vowels among the 5 vowel positions. This is 5! ways.Step 4: Count permutations of the 4 distinct consonants among the 4 consonant positions. This is 4! ways.Step 5: Multiply the independent counts for vowels and consonants. Total arrangements = 5! * 4!.Step 6: Compare with the given options. The expression 5! * 4! matches option written as 4! x 5!. Verification / Alternative check:An alternative way to think about the problem is to imagine two separate slots: one row of vowel slots and one row of consonant slots. First, arrange the 5 vowels in all possible orders, which gives 5! permutations. Then separately arrange the 4 consonants, which gives 4! permutations. Because every full word must be obtained by combining one vowel arrangement with one consonant arrangement, again the total count is 5! * 4!. There is no other restriction, so this confirms the result. Why Other Options Are Wrong: 4! x 4!: This would imply that there are only 4 vowels and 4 consonants, which is incorrect because there are 5 vowels in EDUCATION. 5! x 5!: This assumes 5 vowels and 5 consonants, but the word has only 4 consonants, so this overcounts arrangements. 3! x 4!: This assumes only 3 vowel positions being permuted, which does not match the actual structure of the word. Common Pitfalls:Students often forget to classify letters correctly and may miscount vowels or consonants. Another common mistake is to think that the pattern of actual letters must remain fixed, instead of the pattern of vowel and consonant positions. Yet another error is to simply take 9! permutations without considering the restriction on positions. Some learners also confuse 5! + 4! with 5! * 4!, but in problems where two independent choices are made, multiplication is the correct rule of counting. Final Answer:The total number of valid rearrangements is 4! x 5!, which corresponds to 5! vowel permutations times 4! consonant permutations.

Question 10

A decision committee of 5 members is to be formed from a group of 4 actors, 3 directors and 2 producers. In how many distinct ways can we form a committee consisting of exactly 2 actors, 2 directors and 1 producer?

Accepted Answer

Introduction / Context:This question checks understanding of combinations in selection problems. We are selecting a fixed size committee from different categories of people: actors, directors and producers. The composition of the committee is restricted, which makes it a standard application of combinations where order does not matter but category counts do matter. Given Data / Assumptions: There are 4 actors, 3 directors and 2 producers in total. A committee of exactly 5 members has to be formed. The committee must contain exactly 2 actors, exactly 2 directors and exactly 1 producer. The order in which members are selected does not matter, only the group composition matters. Concept / Approach:We use combinations because we are choosing people without regard to order. For each category, we count the number of ways to choose the required number of people using nCr. Then, since the selections from actors, directors and producers are independent, the total number of committees equals the product of the combination counts from each category. This is a typical multiplication principle application in combinatorics. Step-by-Step Solution:Step 1: Choose 2 actors from 4 actors. The number of ways is 4C2.Step 2: Compute 4C2 = 4! / (2! * 2!) = 6.Step 3: Choose 2 directors from 3 directors. The number of ways is 3C2.Step 4: Compute 3C2 = 3! / (2! * 1!) = 3.Step 5: Choose 1 producer from 2 producers. The number of ways is 2C1 = 2.Step 6: Multiply the independent choices. Total committees = 4C2 * 3C2 * 2C1 = 6 * 3 * 2 = 36. Verification / Alternative check:An alternative is to think in terms of listing small illustrative subsets. For instance, suppose actors are A1 to A4, directors are D1 to D3 and producers are P1, P2. Any valid committee must contain 2 labels from A set, 2 from D set and 1 from P set. If you start listing a few such committees and count the number of choices at each stage, you will see that for each pair of actors and directors, there are exactly 2 choices of producers. That is, for each of the 6 * 3 = 18 actor director combinations, you can pair P1 or P2, giving 18 * 2 = 36 committees. This matches the earlier calculation. Why Other Options Are Wrong: 18: This counts only 4C2 * 3C2 and ignores the choices for producers. 24: This could arise from a miscalculation, such as 4C2 * 3C1 * 2C1, which uses the wrong director count. 32: This may come from misusing powers of 2 or guessing rather than applying combinations. Common Pitfalls:A frequent error is to treat each committee as an ordered arrangement and use permutations instead of combinations. Another mistake is to forget that selections from each group are independent, leading to incorrect addition instead of multiplication of counts. Some students also misread the requirement and pick the wrong number of directors or actors, which changes the combination values. Carefully mapping each phrase of the question to a combination calculation is essential. Final Answer:The total number of different committees that can be formed is 36 distinct ways.

Question 11

A polygon has exactly 44 diagonals. Using the relationship between the number of sides of a polygon and its diagonals, how many sides does this polygon have?

Accepted Answer

Introduction / Context:This problem tests the formula connecting the number of sides of a polygon and the number of diagonals it has. It is a very common question in geometry based aptitude tests and helps reinforce the idea of counting segments between vertices and excluding edges of the polygon. Given Data / Assumptions: The polygon is simple and has n sides. The total number of diagonals in this polygon is 44. A diagonal is a line segment connecting two non adjacent vertices of the polygon. We assume n is at least 3 because a polygon must have at least 3 sides. Concept / Approach:The formula for the number of diagonals of an n sided polygon is n(n minus 3) / 2. This comes from counting all possible line segments between pairs of vertices and then subtracting the sides of the polygon. We set this formula equal to 44 and solve the resulting quadratic equation to find n. Only the positive integer solution that is meaningful for polygons is accepted. Step-by-Step Solution:Step 1: Use the diagonal formula: number of diagonals = n(n - 3) / 2.Step 2: Set n(n - 3) / 2 equal to 44, so n(n - 3) / 2 = 44.Step 3: Multiply both sides by 2 to remove the denominator: n(n - 3) = 88.Step 4: Expand: n^2 - 3n - 88 = 0.Step 5: Solve the quadratic equation n^2 - 3n - 88 = 0 using factorisation or the quadratic formula. The discriminant is 9 + 352 = 361, whose square root is 19.Step 6: Compute n = (3 ± 19) / 2, so n = (3 + 19) / 2 = 22 / 2 = 11 or n = (3 - 19) / 2 = -16 / 2 = -8. Discard the negative value as invalid for a number of sides. Verification / Alternative check:You can directly verify the result by substituting n = 11 into the diagonals formula. For an 11 sided polygon, diagonals = 11(11 - 3) / 2 = 11 * 8 / 2 = 88 / 2 = 44. This matches the given condition exactly, so n = 11 is correct. Trying any other option, such as 9, would give 9(9 - 3) / 2 = 9 * 6 / 2 = 54 / 2 = 27 diagonals, which does not match the required value. Why Other Options Are Wrong: 13: Gives diagonals = 13(10) / 2 = 130 / 2 = 65, which is more than 44. 9: Gives diagonals = 9(6) / 2 = 54 / 2 = 27, less than 44. 7: Gives diagonals = 7(4) / 2 = 28 / 2 = 14, much less than 44. Common Pitfalls:Students sometimes confuse the diagonal formula with nC2 or make sign errors when forming the quadratic equation. Another frequent error is to forget to discard the negative root, even though a polygon cannot have a negative number of sides. Some may also misinterpret the problem and try to count edges instead of diagonals. Carefully applying the correct formula and checking results against the given diagonal count avoids these mistakes. Final Answer:The polygon has 11 sides, because an 11 sided polygon has exactly 44 diagonals.

Question 12

Using the digits 2, 3, 4, 5 and 6 without repetition, how many distinct numbers greater than 4000 can be formed, considering both 4 digit and 5 digit numbers?

Accepted Answer

Introduction / Context:This question examines counting of numbers formed from given digits without repetition, under a constraint that the overall number must be greater than 4000. It combines ideas of permutations with positional constraints on the leading digit and requires careful separation of cases for 4 digit and 5 digit numbers. Given Data / Assumptions: Available digits are 2, 3, 4, 5 and 6. No digit may be repeated in any number. We are interested in all numbers greater than 4000. Both 4 digit and 5 digit numbers are allowed. No leading zero issue arises because 0 is not among the digits. Concept / Approach:We count separately the number of valid 4 digit numbers greater than 4000 and the number of valid 5 digit numbers. For 4 digit numbers, we enforce a restriction on the thousands place and then arrange remaining digits. For 5 digit numbers formed with all five digits without repetition, every such number is automatically greater than 4000 because the smallest possible leading digit is 2 which still produces a 5 digit number far above 4000. Step-by-Step Solution:Step 1: Count 4 digit numbers greater than 4000. The thousands digit must be 4, 5 or 6.Step 2: If the leading digit is fixed as 4, the remaining three places can be filled by any permutation of the remaining 4 digits (2,3,5,6), which gives 4P3 = 4 * 3 * 2 = 24 numbers.Step 3: Similarly, if the leading digit is 5, the remaining three places again have 4P3 = 24 possibilities.Step 4: If the leading digit is 6, the remaining three places again have 24 possibilities.Step 5: Total 4 digit numbers greater than 4000 = 24 + 24 + 24 = 72.Step 6: Now count 5 digit numbers. Using all five digits without repetition gives 5! = 120 distinct 5 digit numbers, and every one of them is greater than 4000.Step 7: Total valid numbers = 72 (4 digit) + 120 (5 digit) = 192. Verification / Alternative check:An alternative quick check is to list a smaller example with fewer digits and see the pattern. For instance, if digits were 3, 4, 5 only, you could test how many numbers above a threshold are formed and observe that splitting into cases based on the leading digit gives the correct count. Applying the same logic with five digits matches the formula based approach above. Also, reviewing the counts 72 and 120 separately and confirming that no case is double counted or missed builds confidence in the result. Why Other Options Are Wrong: 120: This counts only the 5 digit permutations and ignores the 4 digit numbers greater than 4000. 256 and 244: These numbers are larger than the correct count and likely come from incorrect use of powers such as 4^4 or from including numbers with repetition, which is not allowed. Common Pitfalls:Some learners forget that 5 digit numbers automatically satisfy the greater than 4000 condition here and therefore undercount. Others accidentally allow repetition of digits or forget to restrict the thousands digit for 4 digit numbers. Another common mistake is to treat 4 digit and 5 digit cases together without careful separation, which can lead to double counting or miscounting. Always split into clear cases when constraints differ between number lengths. Final Answer:The total number of distinct numbers greater than 4000 that can be formed is 192.

Question 13

How many different 4 digit numbers greater than 6000 can be formed using the digits 0, 2, 4 and 6, if repetition of digits is allowed in any position?

Accepted Answer

Introduction / Context:This problem is about constructing 4 digit numbers from a small set of digits, with repetition permitted, and subject to the condition that the final number must be greater than 6000. It checks understanding of place value, especially the restriction on the leading digit, and the use of simple counting with repetition allowed. Given Data / Assumptions: Available digits are 0, 2, 4 and 6. Numbers formed must be 4 digit numbers. Digits may repeat in the same number. The overall number must be greater than 6000. 4 digit numbers cannot start with 0. Concept / Approach:The key idea is to treat each place in the 4 digit number as a slot. The thousands place must be chosen carefully so that the resulting number exceeds 6000. Once the thousands digit is fixed, the remaining positions can be filled freely using the allowed digits because repetition is permitted. We use a simple multiplication principle for counting such arrangements. Step-by-Step Solution:Step 1: For a 4 digit number to be greater than 6000 using digits 0, 2, 4 and 6, the thousands digit must be 6. If it were 0, 2 or 4 the number would be less than 6000.Step 2: So the thousands digit has exactly 1 choice, which is 6.Step 3: The remaining three positions are the hundreds, tens and units places. Each of these can be filled with any of the 4 digits 0, 2, 4 or 6, because repetition is allowed.Step 4: Therefore the hundreds place has 4 choices, the tens place has 4 choices and the units place has 4 choices.Step 5: Using the multiplication principle, total numbers = 1 * 4 * 4 * 4 = 4^3 = 64. Verification / Alternative check: Why Other Options Are Wrong: 63 and 62: These are close to 64 but suggest that the student has subtracted one or two numbers incorrectly, possibly due to mistaken exclusion of numbers like 6000 or confusion about the leading digit condition. 60: This undercounts by not allowing all combinations of the inner digits or by assuming that 0 cannot appear in positions other than the thousands place, which is incorrect. Common Pitfalls:One common error is to allow the thousands digit to be 0, 2 or 4, which would produce numbers less than or equal to 6000 and violate the condition. Another frequent mistake is to think that 0 cannot appear in any position, whereas the restriction only applies to the leading digit in a positive integer. Learners also sometimes forget that repetition is allowed, and they incorrectly use permutations without repetition. Final Answer:The number of different 4 digit numbers greater than 6000 that can be formed is 64.

Question 14

In how many different ways can 21 identical balls be distributed among 3 persons such that each person receives at least 5 balls?

Accepted Answer

Introduction / Context:This question involves distributing identical objects among distinct persons subject to a minimum requirement. It is a standard stars and bars type problem from permutations and combinations, where the objects are identical balls and the people are distinct recipients. The minimum condition makes the problem slightly more interesting than a basic non negative integer solution count. Given Data / Assumptions: There are 21 balls, all identical. There are 3 persons, all distinct. Each person must receive at least 5 balls. No upper limit is given besides the total of 21 balls. Concept / Approach:The usual method is to convert the at least 5 constraint into a non negative constraint by a simple substitution. If each person must receive at least 5 balls, we can give 5 balls to each person first, and then distribute the remaining balls without a lower bound. After this transformation, the problem reduces to finding the number of non negative integer solutions of a linear equation, which is solved with the stars and bars formula. Step-by-Step Solution:Step 1: Let x1, x2 and x3 be the number of balls received by the three persons.Step 2: We have x1 + x2 + x3 = 21, with each xi greater than or equal to 5.Step 3: Define new variables y1 = x1 - 5, y2 = x2 - 5, y3 = x3 - 5. Each yi is now greater than or equal to 0.Step 4: Substituting, we get (y1 + 5) + (y2 + 5) + (y3 + 5) = 21, so y1 + y2 + y3 + 15 = 21.Step 5: Simplify to get y1 + y2 + y3 = 6, with yi greater than or equal to 0.Step 6: The number of non negative integer solutions of y1 + y2 + y3 = 6 is given by (6 + 3 - 1)C(3 - 1) = 8C2.Step 7: Compute 8C2 = 8 * 7 / (2 * 1) = 28. Verification / Alternative check:You can verify the count by listing some sample solutions. For example, when y1 = 0, y2 + y3 = 6, which has 7 solutions (including 0,6 and 6,0). When y1 = 1, there are 6 solutions, and so on. Summing these manually produces 28 solutions and matches the stars and bars formula. This confirms that there are 28 valid distributions of the original 21 balls with each person getting at least 5. Why Other Options Are Wrong: 14 and 21: These numbers can appear if someone forgets to convert to the non negative form correctly or misapplies the combination formula. 7: This is far too small and could result from only counting distributions with one person fixed and ignoring others. Common Pitfalls:A typical mistake is to directly apply the formula (21 + 3 - 1)C(3 - 1) without adjusting for the minimum of 5, which would ignore the lower bound requirement. Another mistake is to treat the balls as distinct, which is incorrect for this problem. Also, some learners only distribute the remaining balls to two people instead of three after the transformation, which changes the equation and the answer. Final Answer:The balls can be distributed among the three persons in 28 different ways.

Question 15

If nC10 is equal to nC12 for some positive integer n, then what is the value of n that satisfies this binomial coefficient equality?

Accepted Answer

Introduction / Context:This question tests knowledge of binomial coefficients and symmetry properties of nCr. When two binomial coefficients with the same n but different r values are equal, there is a specific relationship between the r values and n. Understanding this relationship is important in many combinatorial proofs and problems. Given Data / Assumptions: We are given that nC10 = nC12 for some integer n. We assume n is a positive integer and at least as large as the larger r, so n is at least 12. We want to determine the value of n that makes these two coefficients equal. Concept / Approach:A key property of binomial coefficients is the symmetry nCr = nC(n - r). If two coefficients with same n are equal and r values are different, then typically one of them equals n minus the other. In other words, if nCr1 = nCr2 and r1 is not equal to r2, we usually have r1 + r2 = n. We apply this concept directly here with r1 = 10 and r2 = 12. Step-by-Step Solution:Step 1: Use the property: if nCr1 = nCr2 and r1 is not equal to r2, then r1 + r2 = n.Step 2: Here r1 = 10 and r2 = 12.Step 3: Add the r values: 10 + 12 = 22.Step 4: Therefore, n must be 22 to satisfy nC10 = nC12.Step 5: Optionally verify using symmetry: for n = 22, nC10 = 22C10 and nC12 = 22C(22 - 10) = 22C12. Because 10 and 12 sum to 22, the symmetry property guarantees equality. Verification / Alternative check:As an algebraic check, you could write nC10 and nC12 using factorials and equate them. After cancelling common factorial factors, the equation simplifies and leads to n = 22. This process is longer but matches the quick symmetry based method and confirms that there is a unique positive integer solution for n in this context. Why Other Options Are Wrong: 10 and 12: These are values of r, not n. Choosing them confuses the role of n and r in the binomial coefficient. 24: This would make r1 + r2 less than n, and 24C10 would equal 24C14, not 24C12, so the equality would not hold. Common Pitfalls:A common mistake is to try to expand the factorials fully, which is time consuming and error prone. Another is to think that the equality holds for any n greater than or equal to 12, which is not true. Some students also incorrectly apply properties like nCr = nC(r + 1), which do not hold. Remembering the correct symmetry relation nCr = nC(n - r) and using r1 + r2 = n is the clean way to handle such problems. Final Answer:The value of n for which nC10 = nC12 is 22.

Question 16

Four ladies A, B, C and D and four gentlemen E, F, G and H are sitting around a circular table facing each other. No two ladies sit together and no two gentlemen sit together. C, who is between G and E, is facing D. F is between D and A and is facin…

Accepted Answer

Introduction / Context:This is a typical circular seating arrangement and logical reasoning question. It tests the ability to interpret multi condition statements, place people around a circle and then deduce relative positions like neighbours and who sits opposite whom. Such problems are common in competitive exams and require a systematic step by step approach. Given Data / Assumptions: There are four ladies A, B, C, D and four gentlemen E, F, G, H. All are sitting around a circular table facing the centre. No two ladies are adjacent and no two gentlemen are adjacent, so seats alternate lady and gentleman. C sits between G and E and faces D. F is between D and A and faces G. H sits immediately to the right of B. We consider right as clockwise direction when people face the centre. Concept / Approach:We use a structured reasoning method. First, exploit symmetry by fixing one persons position to reduce rotational ambiguity. Then place others using the constraints one by one. Because genders must alternate, we know pattern of lady and gentleman seats around the circle. Using the between and facing conditions we can identify exact seats. Finally, we read off neighbours and opposite seats for the required persons. Step-by-Step Solution:Step 1: Because ladies and gentlemen alternate, assume seats 1, 3, 5, 7 are ladies and seats 2, 4, 6, 8 are gentlemen.Step 2: Fix C at seat 1 without loss of generality (circular symmetry). Then D, who is facing C, must be at seat 5.Step 3: C is between G and E. So seats 8 and 2, the neighbours of C, must be G and E in some order.Step 4: F is between D and A and faces G. F is a gentleman, so F occupies some gentleman seat that is opposite G.Step 5: Try G at seat 2 and E at seat 8. Then the opposite of seat 2 is seat 6, so F must be at seat 6. F is between D and A, so the neighbours of F, seats 5 and 7, must be D and A. Since seat 5 is already D, seat 7 must be A.Step 6: The remaining lady B must be at seat 3. The only unused gentleman seat is 4, so H must be at seat 4.Step 7: Check H is to the right of B. From B at seat 3, the immediate right (clockwise) seat is 4, which is H, so this arrangement satisfies all conditions.Step 8: Now read answers. Neighbours of B (seat 3) are G at seat 2 and H at seat 4. E is at seat 8, which is opposite seat 4, so E faces H. Verification / Alternative check:To be sure this is unique, try the other initial possibility where G is at seat 8 and E is at seat 2. In that case, F would need to be opposite G at seat 4. For F to be between D and A, the neighbours of seat 4 (seats 3 and 5) must be D and A, but seat 5 is already D, so A would be at seat 3 and B at seat 7. Then the right of B at seat 7 is seat 8, which would be G instead of H, violating the condition that H is to the right of B. Hence the previous arrangement is the only valid one, confirming that B has neighbours G and H and E faces H. Why Other Options Are Wrong: F & H , B: This would require B to be opposite E and for F to be adjacent to B, which is not consistent with the derived seating pattern. E & F , F: This incorrectly places F both as a neighbour and as the person facing E, which cannot happen with alternating genders and the given conditions. E & H , G: This would have E and H beside B and E facing G, which does not match the only consistent arrangement satisfying all statements. Common Pitfalls:People often misinterpret right and left in circular arrangements or forget that everyone is facing the centre. Another major mistake is not respecting the alternation of ladies and gentlemen and placing two of the same gender together. Some also stop after a partial arrangement without checking all conditions, leading to a wrong conclusion. Always verify every given condition once you think you have a solution. Final Answer:The immediate neighbours of B are G and H, and E is facing H.

Question 17

From a basket that contains exactly 3 oranges, 5 apples and 7 mangoes (15 fruits in total), in how many different ways can a selection be made by taking any number of fruits, including possibly all 15, but not taking none?

Accepted Answer

Introduction / Context:This is a combinations with identical objects problem, where the fruits of the same type are indistinguishable. We count selections based on how many of each fruit type are chosen, subject to stock limits and the condition that at least one fruit is selected. It helps practise counting with upper bounds and exclusion of the empty selection. Given Data / Assumptions: There are 3 oranges, 5 apples and 7 mangoes. Fruits of the same type are considered identical for selection purposes. Any number of fruits can be taken, from 1 up to all 15. Selecting no fruit at all is not allowed. Concept / Approach:We count the number of possible triples (o, a, m), where o is the number of oranges chosen, a is the number of apples and m is the number of mangoes. Each variable is bounded between 0 and its stock limit. The total number of such triples, including the case where all three counts are zero, is simply the product of the possible choices for each type. Then we subtract the one disallowed case where no fruit is chosen. Step-by-Step Solution:Step 1: Let o be the number of oranges chosen. Since there are 3 oranges, o can be 0, 1, 2 or 3. This gives 4 possible values.Step 2: Let a be the number of apples chosen. Since there are 5 apples, a can be 0, 1, 2, 3, 4 or 5. This gives 6 possible values.Step 3: Let m be the number of mangoes chosen. Since there are 7 mangoes, m can be any integer from 0 to 7 inclusive, giving 8 possible values.Step 4: Without any restriction on taking at least one fruit, total selections correspond to the number of ordered triples (o, a, m) with these ranges, which is 4 * 6 * 8 = 192.Step 5: One of these triples corresponds to taking no fruit at all, namely (0, 0, 0). This selection is not allowed.Step 6: Therefore, valid selections = 192 - 1 = 191. Verification / Alternative check:A quick way to verify the idea is to look at a simpler example, such as 1 orange and 2 apples. In that small case, you can list all valid selections explicitly and see that the count equals (1 + 1) * (2 + 1) minus 1. Extending this logic to 3, 5 and 7 fruits per type gives the same product minus one pattern used here. No double counting occurs because each choice triple is unique. Why Other Options Are Wrong: 131 and 68: These undercount and likely result from partial enumeration or forgetting that each type can have zero chosen. 3720: This is far too large and suggests someone treated fruits as distinct objects and used permutations, which does not match the problem statement with identical fruits per type. Common Pitfalls:A very common mistake is to assume fruits are distinct and use large permutation or combination counts that are not appropriate. Another error is to forget to subtract the empty selection and accept 192 instead of 191. Some students also misread the limits, such as thinking there are 15 independent choices rather than grouped by fruit type. Always focus on how many of each type you can choose. Final Answer:The number of different possible selections is 191.

Question 18

How many different 4 letter words (not necessarily meaningful) can be formed using the letters of the word MEDITERRANEAN such that the first letter is E and the last letter is R?

Accepted Answer

Introduction / Context:This question combines permutations with repeated letters. The word MEDITERRANEAN has several repeated letters, and we restrict the first and last positions of the 4 letter word. We must count distinct arrangements respecting letter multiplicities and position constraints, a typical challenge in combinatorics. Given Data / Assumptions: The base word is MEDITERRANEAN. Letter counts are: E 3 times, R 2 times, A 2 times, N 2 times, and M, D, I, T once each. We need 4 letter words with first letter fixed as E and last letter fixed as R. Letters cannot be used more times than they appear in the original word. Words are considered distinct based on letter sequence; they need not be meaningful. Concept / Approach:Because the first and last letters are fixed, we only need to choose and arrange the two middle letters. We treat MEDITERRANEAN as a multiset of available letters. After using one E and one R for the first and last positions, we find how many distinct 2 letter sequences can be formed from the remaining letters, respecting their remaining counts. This is a multiset permutation problem of length 2. Step-by-Step Solution:Step 1: Count letters in MEDITERRANEAN. We have E 3, R 2, A 2, N 2 and M, D, I, T 1 each.Step 2: Fix the first letter as E and the last letter as R. After using one E and one R, the remaining counts are: E 2, R 1, A 2, N 2, M 1, D 1, I 1, T 1.Step 3: We must fill the second and third positions using these remaining letters.Step 4: Consider the 8 distinct letter types {E, R, A, N, M, D, I, T}. We count all possible ordered pairs that respect multiplicities.Step 5: If the two middle letters are different, any ordered pair of two different letters from the 8 types is allowed, since only one of each type is used. Number of such ordered pairs is 8 * 7 = 56.Step 6: If the two middle letters are the same, we need letters with at least 2 copies remaining. These are E (2 left), A (2) and N (2), contributing middle pairs EE, AA and NN. That gives 3 additional words.Step 7: Total distinct 4 letter words = 56 + 3 = 59. Verification / Alternative check: Why Other Options Are Wrong: 56: This omits the cases where the two middle letters are the same and counts only distinct letter pairs. 64 and 55: These values do not correspond to any correct splitting into identical and distinct pairs and likely come from miscounting the multiplicity of E, A and N. Common Pitfalls:Students may treat all remaining letters as distinct and simply compute 8^2 = 64, ignoring the upper bounds from the original word. Others count only combinations without considering order, which would undercount. A frequent oversight is forgetting that E, A and N can each appear twice in the middle positions because they still have at least 2 copies left. Final Answer:The number of different 4 letter words with first letter E and last letter R is 59.

Question 19

There are 7 non collinear points in a plane. By joining these points in pairs, how many distinct triangles can be formed?

Accepted Answer

Introduction / Context:This question tests counting of geometric shapes using combinations. When no three points are collinear, every set of three points forms a unique triangle. This leads directly to a combination based formula, which is a standard result in combinatorial geometry. Given Data / Assumptions: There are 7 points in a plane. No three points lie on the same straight line (non collinear condition). Each triangle is determined by 3 distinct points. Triangles are considered distinct based on their vertex sets. Concept / Approach:If no three points are collinear, then any selection of 3 distinct points gives a valid triangle. Therefore, the number of triangles equals the number of combinations of 7 points taken 3 at a time, which is 7C3. We simply compute this combination. Step-by-Step Solution:Step 1: The number of triangles is the number of ways to choose 3 points out of 7.Step 2: Use the combination formula 7C3 = 7! / (3! * 4!).Step 3: Compute 7! = 7 * 6 * 5 * 4 * 3 * 2 * 1.Step 4: Use cancellation: 7! / (3! * 4!) simplifies to (7 * 6 * 5) / (3 * 2 * 1) = 210 / 6.Step 5: Evaluate 210 / 6 = 35. Verification / Alternative check:A quick check is to consider smaller numbers of points, such as 4 non collinear points. In that case, 4C3 = 4 triangles, which you can easily visualise. Extending to 7 points and using the same formula gives 7C3 = 35, which is consistent with this general pattern and uses the same underlying combinatorial reasoning. Why Other Options Are Wrong: 10: This equals 5C2, not 7C3, and undercounts greatly. 30: This might come from an incorrect formula or a calculation slip like 6 * 5 instead of 7C3. 60: This is too large and could result from using permutations instead of combinations. Common Pitfalls:Students sometimes forget the non collinear condition and try to subtract extra cases or worry about degenerate triangles. Another mistake is to use permutations 7P3, which would count ordered triples and not triangles. Always recall that a triangle is determined only by the set of 3 vertices, so combinations are appropriate. Final Answer:The number of distinct triangles that can be formed is 35.

Question 20

In a cricket club there are 6 bowlers and 9 batsmen. In how many different ways can a team of 11 players be selected such that the team contains at least 4 bowlers?

Accepted Answer

Introduction / Context:This question involves combinations with category based constraints. We need to form an 11 player cricket team from a pool of bowlers and batsmen, ensuring that the team has at least 4 bowlers. It is a classic example of selecting with lower bounds on one category. Given Data / Assumptions: Total bowlers available: 6. Total batsmen available: 9. Team size must be exactly 11 players. The team must contain at least 4 bowlers. All players are distinct; order of selection does not matter. Concept / Approach:We break the problem into cases based on the exact number of bowlers in the team: 4, 5 or 6 bowlers (at least 4 means 4 or more, but there are only 6 bowlers available). For each case, we choose the corresponding number of bowlers and batsmen using combinations and then sum the counts from all valid cases. Step-by-Step Solution:Step 1: Case 1 (4 bowlers): Choose 4 bowlers out of 6 and 7 batsmen out of 9. The number of ways is 6C4 * 9C7.Step 2: Compute 6C4 = 6C2 = 15 and 9C7 = 9C2 = 36. Thus, Case 1 gives 15 * 36 = 540 ways.Step 3: Case 2 (5 bowlers): Choose 5 bowlers out of 6 and 6 batsmen out of 9. The number of ways is 6C5 * 9C6.Step 4: Compute 6C5 = 6C1 = 6 and 9C6 = 9C3 = 84. Thus, Case 2 gives 6 * 84 = 504 ways.Step 5: Case 3 (6 bowlers): Choose all 6 bowlers and then 5 batsmen out of 9. The number of ways is 6C6 * 9C5 = 1 * 126 = 126 ways.Step 6: Add the three cases. Total valid teams = 540 + 504 + 126 = 1170. Verification / Alternative check: Why Other Options Are Wrong: 1200 and 720: These values do not match the sum of any correct combination of cases and likely come from incorrect use of combinations or ignoring some cases. 360: This is much too small and might arise if someone only counted one of the valid cases, such as 6C4 * 9C7 but miscomputed it. Common Pitfalls:Students sometimes forget that at least 4 bowlers includes 5 and 6 bowlers, not just exactly 4, leading to undercounting. Another common mistake is to ignore that the total must be 11 and incorrectly pair incorrect numbers of batsmen with a given number of bowlers. It is also easy to mix up when to add and when to multiply counts, so clearly separating cases is essential. Final Answer:The number of different 11 player teams that can be formed with at least 4 bowlers is 1170.

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