Question 1

From a group of 4 people, in how many different ways can we choose a group of 3 people if the order of people in the group does not matter?

Accepted Answer

Introduction / Context: This is a very simple combinations question. We have a small set of people and we must form a smaller group, where the order of selection does not matter. Questions like this introduce the basic concept of combinations C(n,r). Given Data / Assumptions: Total number of people available = 4. We must form a group of exactly 3 people. Each person is distinct. Order of people inside the group is irrelevant. Concept / Approach: The number of ways to choose r objects from n distinct objects when order does not matter is given by C(n,r) = n! / (r! * (n - r)!). Here, n = 4 and r = 3. This formula automatically accounts for the fact that different orders of the same people are not counted separately, which is what we want for a committee or group. Step-by-Step Solution: We need to compute C(4,3). Using the formula: C(4,3) = 4! / (3! * 1!). Compute 4! = 4 * 3 * 2 * 1 = 24. Compute 3! = 3 * 2 * 1 = 6. Now C(4,3) = 24 / (6 * 1) = 24 / 6 = 4. So there are 4 different groups of 3 people that can be selected. Verification / Alternative check: We can list possibilities to confirm. Suppose the four people are A, B, C and D. The possible 3 person groups are: ABC, ABD, ACD and BCD. We can see there are only 4 distinct groups and that this matches the value of C(4,3). This simple listing check is possible because the numbers are very small. Why Other Options Are Wrong: 12: This is 4P3, the number of permutations, and incorrectly treats different orders of the same group as different outcomes. 36: This is even larger and corresponds to overcounting by adding extra factors not justified by the problem. 16: This figure does not come from any standard nCr or nPr value for n = 4 and r = 3. Common Pitfalls: The main pitfall is confusion between permutations and combinations. For groups or committees where order does not matter, combinations must be used. Students sometimes automatically use nPr, which leads to inflated answers. Another issue is not recognising that C(4,3) equals C(4,1), which can sometimes give a quick shortcut. Always check whether position or order is important before choosing the formula. Final Answer: The number of different 3 person groups that can be formed from 4 people is 4.

Question 2

In a standard deck of cards there are four eights and four sevens. How many different 5 card hands can be dealt that consist of exactly three eights and two sevens?

Accepted Answer

Introduction / Context: This question is about counting specific types of card hands using combinations. We are asked to find how many 5 card hands contain exactly three cards of one rank (eight) and exactly two cards of another rank (seven). The key is to recognise that within each rank we are choosing a subset of the available cards. Given Data / Assumptions: Standard deck: 52 playing cards. Number of eights in the deck = 4. Number of sevens in the deck = 4. We must form a hand of exactly 5 cards. Hand must contain exactly three eights and exactly two sevens. Order of cards in a hand does not matter. Concept / Approach: To construct such a hand, we must independently choose which eights appear and which sevens appear. Since there are 4 cards of each rank, the number of ways to choose 3 eights is C(4,3), and the number of ways to choose 2 sevens is C(4,2). Because these choices are independent, we multiply the two combination counts to get the total number of distinct hands fitting the requirement. Step-by-Step Solution: Number of ways to choose 3 eights from 4 eights = C(4,3). Compute C(4,3) = 4! / (3! * 1!) = 4. Number of ways to choose 2 sevens from 4 sevens = C(4,2). Compute C(4,2) = 4! / (2! * 2!) = 6. Total hands with exactly three eights and two sevens = C(4,3) * C(4,2) = 4 * 6. Compute 4 * 6 = 24. Verification / Alternative check: Since a hand must contain exactly 3 + 2 = 5 cards, and there are only two ranks involved, there is no flexibility to include other ranks. Each valid hand is therefore uniquely determined by which specific three eights and which specific two sevens it contains. There is no risk of double counting because we never change the hand structure, only choose specific cards within each group. The multiplication rule thus gives the correct answer. The small final number reflects the strict requirement. Why Other Options Are Wrong: 36: This might come from incorrectly using C(4,3) * C(4,3) or another wrong combination count. 72: This is three times the correct answer, perhaps from mistakenly multiplying by an extra factor such as 3!. 16: This is too small and suggests simply 4 * 4, which would be wrong because we must choose combinations, not count ordered pairs. Common Pitfalls: Typical mistakes include treating eights and sevens as if any number of each can appear, which leads to unwanted hands with different compositions, or using permutations instead of combinations. Some students also forget the condition "exactly three" eights and "exactly two" sevens and accidentally include cases with four eights and one seven or similar variations. Clearly restating the required counts before computing combinations helps avoid these errors. Final Answer: The number of different 5 card hands that contain exactly three eights and two sevens is 24.

Question 3

In how many different ways can the letters of the word "ABACUS" be rearranged so that all the vowels appear together as a single block?

Accepted Answer

Introduction / Context: This problem is another example of arranging letters with some of them treated as a group. The word "ABACUS" contains repeating letters and several vowels. We must count permutations where the vowels are together, which requires grouping them into a block and then handling repetition both within that block and in the overall arrangement. Given Data / Assumptions: Word: ABACUS. Total letters = 6. Vowels: A, A, U (3 vowels with A repeated twice). Consonants: B, C, S (3 distinct consonants). All vowels must appear together as one contiguous block in every valid arrangement. All letters of the word are used in each arrangement. Concept / Approach: We use a block method with two stages. First, treat the group of vowels A, A and U as a single super letter or block. This block then joins the consonants B, C and S to form fewer effective objects. We count permutations of these objects. Second, we count the internal permutations of the vowels within the block, again compensating for the repeated A. The final answer is the product of these two counts. Step-by-Step Solution: Step 1: Treat the vowels A, A, U as a single block V. Now the objects are V, B, C and S. That means we have 4 distinct objects. Number of ways to arrange these 4 objects = 4! = 24. Step 2: Arrange the vowels inside the block V. The letters inside the block are A, A and U. Number of internal arrangements = 3! / 2! (since A repeats twice). Compute 3! = 6; divide by 2 to get 3. Total valid arrangements = (ways to arrange V, B, C, S) * (ways to arrange A, A, U). Total = 24 * 3 = 72. Verification / Alternative check: Check that the result is less than the total number of permutations of ABACUS without restriction. Without any grouping, total distinct permutations would be 6! / 2! because only A repeats. That equals 720 / 2 = 360. Our answer 72 is exactly one fifth of 360, which is reasonable since we force three letters to stick together. The method separates the external arrangement and internal arrangement correctly and respects the repeated letters, so we can be confident in 72 as the correct value. Why Other Options Are Wrong: 12: This equals 4! / 2 and incorrectly treats the block or the repetition. 3: This is only the number of internal vowel arrangements and ignores the outer positions. 36: This is half of the correct value and usually comes from missing a factor of 2 somewhere in the counting. Common Pitfalls: Common errors include forgetting that the two A letters are identical, so dividing by 2! is necessary, or failing to treat the vowel group as a single block before applying the factorial count. Some learners also accidentally treat the block plus consonants as having fewer or more than 4 objects. Keeping the two stage structure clear and double checking the list of objects in each stage is the best way to avoid mistakes. Final Answer: The number of different arrangements of "ABACUS" in which all vowels appear together is 72.

Question 4

The word "PROBLEM" has 7 distinct letters. In how many different ways can all the letters be rearranged to form 7 letter words if no letter is repeated?

Accepted Answer

Introduction / Context: This is a pure permutation question involving a word with all distinct letters. We must count the number of ways to rearrange all letters of the word, which directly applies the factorial rule for permutations when there are no repetitions. Given Data / Assumptions: Word: PROBLEM. Letters: P, R, O, B, L, E, M, all distinct. Total letters = 7. All 7 letters must be used in each arrangement. No letter is repeated within an arrangement. Different orders of the same letters are considered different words. Concept / Approach: For n distinct objects arranged in a row with all used exactly once, the number of permutations is simply n!. No adjustment for repetition is needed here because every letter appears exactly once. Hence we compute 7! for this problem. Step-by-Step Solution: Total letters to arrange = 7. Number of permutations of 7 distinct letters = 7!. Compute 7! = 7 * 6 * 5 * 4 * 3 * 2 * 1. First, 7 * 6 = 42 and 5 * 4 = 20. Then 42 * 20 = 840. Next multiply by 3 * 2 * 1 = 6. Finally, 840 * 6 = 5040. Verification / Alternative check: We can think position by position. There are 7 choices for the first position, 6 for the second, 5 for the third and so on until there is only 1 choice left for the last position. The total number is therefore 7 * 6 * 5 * 4 * 3 * 2 * 1, which again equals 7!. This repeated reasoning confirms that 5040 is the correct count of different 7 letter arrangements. Why Other Options Are Wrong: 49: This is 7^2 and could come from misinterpreting the structure of the word or using an incorrect formula. 823543: This equals 7^7 and would be the count if each of the 7 positions could independently be any of 7 letters with repetition allowed, which is not the case. 343: This equals 7^3 and has no connection to the correct permutation logic here. Common Pitfalls: The main error learners make is mixing up permutations with combinations or using exponent rules like 7^n without thinking about whether repetition is allowed. Another issue is forgetting that order matters when forming words or sequences. When every letter is distinct and all must be used once, the factorial rule n! is a simple and reliable tool. Final Answer: The number of different 7 letter words that can be formed from "PROBLEM" is 5040.

Question 5

In the Service Club there are 14 juniors and 23 seniors. The club wants to send 4 representatives to a state conference, with exactly 2 juniors and 2 seniors. How many different groups of representatives are possible?

Accepted Answer

Introduction / Context: This problem is about forming a small team from two categories of members while respecting a fixed composition. Such questions test understanding of combinations and how to multiply combination counts from different groups when the selections are independent. Given Data / Assumptions: Number of juniors in the club = 14. Number of seniors in the club = 23. Total people available = 14 + 23 = 37. We must choose exactly 4 representatives. Required composition: 2 juniors and 2 seniors. Order of representatives in the group does not matter. Concept / Approach: The selection can be broken into two independent choices: choosing juniors and choosing seniors. For each group we use combinations because we only care about which people are chosen, not the order in which they are picked. The total number of valid groups is then the product of the number of ways to choose juniors and the number of ways to choose seniors. Step-by-Step Solution: Number of ways to choose 2 juniors from 14 juniors = C(14,2). Compute C(14,2) = 14! / (2! * 12!) = (14 * 13) / 2 = 91. Number of ways to choose 2 seniors from 23 seniors = C(23,2). Compute C(23,2) = 23! / (2! * 21!) = (23 * 22) / 2 = 253. Total number of valid representative groups = C(14,2) * C(23,2) = 91 * 253. Compute 91 * 253: first 91 * 200 = 18200, then 91 * 50 = 4550 and 91 * 3 = 273. Add 18200 + 4550 + 273 = 23023. Verification / Alternative check: The result should be substantially larger than either 91 or 253, because we are combining choices from two groups. An approximate product 100 * 250 would give about 25000, so 23023 is a reasonable exact value. There is no overlap between the decisions about juniors and seniors, and each valid combination of juniors and seniors defines a unique group of 4 representatives, so the multiplication method is correct. Why Other Options Are Wrong: 23024: Off by 1 from the correct answer, usually from a small arithmetic slip. 24023: Larger than the true product of 91 and 253, so it cannot be correct. 25690: Much larger and does not match any natural combination product from the given numbers. Common Pitfalls: A typical mistake is to use C(37,4), which counts all 4 person committees without regard to the required mix of juniors and seniors. Another error is to miscompute C(14,2) or C(23,2) by forgetting to divide by 2 or by mishandling the multiplication. Writing out the combination formulas clearly and computing the products in stages helps to avoid these errors. Final Answer: The number of different 4 person groups containing exactly 2 juniors and 2 seniors is 23023.

Question 6

From a standard deck of 52 playing cards, a 5 card hand is dealt. In how many distinct ways can the hand contain at least 2 queens?

Accepted Answer

Introduction / Context: This question asks for the number of 5 card hands that contain at least two queens. It is a combinatorics problem involving casework on how many queens appear. We need to carefully consider separate situations for exactly 2 queens, exactly 3 queens and all 4 queens, and then add the counts. Given Data / Assumptions: Standard deck: 52 playing cards. Total queens in the deck = 4. Hand size = 5 cards. We need hands with at least 2 queens, meaning 2, 3 or 4 queens are allowed. Order of cards in the hand does not matter. Concept / Approach: We use combinations for hands. For each allowed number of queens k (where k = 2, 3 or 4), we choose which k queens appear using C(4,k), and then fill the remaining 5 - k places with non queen cards chosen from the 48 remaining cards. The total count is the sum over these three cases. This approach avoids double counting, because each hand has a unique number of queens. Step-by-Step Solution: Case 1: Exactly 2 queens in the hand. Choose 2 queens from 4: C(4,2) = 6. Remaining cards needed = 5 - 2 = 3 non queens from 48 others: C(48,3). Case 1 count = C(4,2) * C(48,3). Case 2: Exactly 3 queens in the hand. Choose 3 queens from 4: C(4,3) = 4. Remaining cards needed = 5 - 3 = 2 non queens from 48 others: C(48,2). Case 2 count = C(4,3) * C(48,2). Case 3: All 4 queens in the hand. Choose all 4 queens: C(4,4) = 1. Remaining one card from 48 non queens: C(48,1) = 48. Case 3 count = C(4,4) * C(48,1) = 48. Now compute the values: C(48,3) = 48 * 47 * 46 / (3 * 2 * 1) = 17296. So Case 1 count = 6 * 17296 = 103776. C(48,2) = 48 * 47 / 2 = 1128. So Case 2 count = 4 * 1128 = 4512. Case 3 count = 48. Total hands with at least 2 queens = 103776 + 4512 + 48 = 108336. Verification / Alternative check: We can compare with the total number of 5 card hands, which is C(52,5) = 2598960. The required hands (108336) form a small but plausible fraction of all possible hands. Also, the dominant contribution is from the case of exactly 2 queens, which intuitively should be the most common among hands with at least two queens. The arithmetic for each combination has been broken into small steps to minimise mistakes and the final sum has been verified step by step. Why Other Options Are Wrong: 103336: Slightly smaller than the correct value and likely comes from counting only the 2 queen case and then adding an incorrect figure for the higher cases. 120000: Too large for this specific requirement and not equal to the precise combination sum. 108333: Very close but not exact; it indicates a small addition or subtraction error in combining the cases. Common Pitfalls: Students sometimes misread "at least 2 queens" as "exactly 2 queens" and ignore the 3 and 4 queen hands. Others double count by not splitting the problem into disjoint cases clearly. It is also easy to make arithmetic mistakes when computing large combinations, so writing numerators and denominators explicitly and simplifying carefully is important. Always ensure that your final answer is the sum of all valid, non overlapping cases. Final Answer: The number of 5 card hands that contain at least two queens is 108336.

Question 7

A student council of 5 members is to be formed from a pool of 6 boys and 8 girls. If Jason is one of the boys in the pool and must be included on the council, how many different councils can be formed?

Accepted Answer

Introduction / Context: This question is about forming a committee with the extra condition that a specific person must be included. Such problems are common in combinatorics, and they help illustrate how to fix one member and then count combinations for the remaining positions. Given Data / Assumptions: Total boys in the pool = 6 (including Jason). Total girls in the pool = 8. Total people in the pool = 14. Council size = 5 members. Jason must be on the council. Order of members within the council does not matter. Concept / Approach: Because Jason is required to be on the council, we first fix him as a member. After including Jason, we only need to choose the remaining 4 members from the remaining 13 people. Since the council is simply a group with no internal order, we use combinations C(n,r) to count the ways to choose these 4 additional members. Step-by-Step Solution: Total people in the pool = 14. Jason is always included, so he occupies one of the 5 council seats. Remaining people available = 14 - 1 = 13. Remaining seats to fill on the council = 5 - 1 = 4. Number of ways to choose 4 people from the remaining 13 = C(13,4). Compute C(13,4) = 13! / (4! * 9!). In product form, C(13,4) = (13 * 12 * 11 * 10) / (4 * 3 * 2 * 1). Compute numerator: 13 * 12 = 156, 11 * 10 = 110, then 156 * 110 = 17160. Compute denominator: 4 * 3 * 2 * 1 = 24. Now 17160 / 24 = 715. Verification / Alternative check: We can observe that C(13,4) should be a mid sized number since 13 is not very large. Also, C(13,4) equals C(13,9), reflecting symmetry in combinations. Known values around this range include C(10,4) = 210 and C(12,4) = 495, so 715 for C(13,4) fits the pattern of increasing values. Since Jason is fixed and we use all other choices only once, there is no double counting. Why Other Options Are Wrong: 725: Slightly larger than C(13,4) and not equal to any standard combination from these numbers. 419: Too small; it misses a large portion of valid councils. 341: Also far from C(13,4) and does not correspond to any correct counting method here. Common Pitfalls: One common mistake is to forget to fix Jason first and instead compute C(14,5), which counts councils where Jason may or may not be present. Another is to think that Jason's fixed presence adds an extra factor like 5, which leads to overcounting. The safe approach is to treat Jason as already included and then apply the combinations formula to the remaining positions. Final Answer: The number of different student councils of 5 members that include Jason is 715.

Question 8

How many distinct 5-letter arrangements can be formed from the letters of the word TRIGONAL if each arrangement uses exactly two of the vowels (A, I, O) and exactly three of the consonants, with no letter repeated?

Accepted Answer

Introduction / Context: This problem tests core ideas from permutations and combinations. The word TRIGONAL has several distinct letters, including both vowels and consonants. We are asked to build 5-letter arrangements that obey two simultaneous constraints: exactly two vowels must be used and exactly three consonants must be used, with each letter used at most once. Such questions are common in aptitude exams because they combine selection (combinations) with arrangement (permutations), and they require careful counting of how many choices are possible at each stage. Given Data / Assumptions: The word TRIGONAL has 8 distinct letters: T, R, I, G, O, N, A, L. Vowels in the word: A, I, O (3 vowels in total). Consonants in the word: T, R, G, N, L (5 consonants in total). We must form 5-letter arrangements. Exactly 2 vowels and exactly 3 consonants must be used. No letter may be repeated within an arrangement. Concept / Approach: The solution proceeds in two stages. First, we choose which actual letters will be included in each 5-letter arrangement. Second, we count how many different ways these chosen letters can be permuted. Choosing letters is a combinations problem because we are selecting subsets of vowels and consonants. Arranging the chosen letters is a permutations problem because order matters in forming different 5-letter words. The total number of valid arrangements is the product of the number of ways to choose the letters and the number of ways to arrange them. Step-by-Step Solution: Step 1: Choose 2 vowels out of the 3 vowels A, I, O. Number of ways to choose the vowels = C(3, 2). C(3, 2) = 3. Step 2: Choose 3 consonants out of the 5 consonants T, R, G, N, L. Number of ways to choose consonants = C(5, 3). C(5, 3) = 10. Step 3: For every chosen set of 5 letters (2 vowels + 3 consonants), arrange them in all possible orders. Number of arrangements of 5 distinct letters = 5! = 120. Step 4: Multiply all stages to get the total number of valid 5-letter arrangements. Total arrangements = C(3, 2) * C(5, 3) * 5!. Total arrangements = 3 * 10 * 120 = 3600. Verification / Alternative check: Another way to verify is to list the counting logic verbally: there are 3 ways to choose which 2 vowels appear, 10 ways to choose which 3 consonants appear, and for each specific choice of 5 letters, there are 120 different orders. Because all letters are distinct, no overcounting occurs. The multiplication principle therefore applies correctly, and the final number 3600 is consistent with standard combinatorial reasoning. Why Other Options Are Wrong: Option 6300 overcounts and does not correspond to any correct combination and permutation breakdown. Option 2400 is too small; it would arise only if either the number of vowel choices or consonant choices was miscounted. Option 7200 doubles the correct answer, which usually comes from mistakenly allowing 6 letters instead of 5 or from counting some step twice. Option 1800 is exactly half of the correct total, which could result from forgetting the contribution of either vowel choices or consonant choices. Common Pitfalls: A common mistake is to choose the total 5 letters directly from all 8 letters and then try to adjust for the vowel and consonant constraints, which is more complicated and error prone. Another typical error is to forget that the letters are distinct, which might lead to wrong use of factorials or division by extra symmetry factors. Some learners also accidentally use permutations instead of combinations when selecting vowels and consonants, or they forget to multiply by 5! at the end. Carefully separating the selection stage from the arrangement stage avoids these issues. Final Answer: Thus, the total number of distinct 5-letter arrangements of TRIGONAL that contain exactly two vowels and three consonants is 3600.

Question 9

A research team of 6 people is to be formed from 10 chemists, 5 politicians, 8 economists and 15 biologists. In how many different ways can such a team be chosen if the team must contain at least 5 chemists?

Accepted Answer

Introduction / Context: This problem is a classic example of forming a committee with a restriction on the composition of its members. We are given several different categories of specialists and are asked to count the number of ways to form a research team of fixed size when at least a minimum number of chemists must be included. Such questions check the ability to break a counting problem into cases and use combinations correctly while satisfying a constraint like “at least k of a particular type”. Given Data / Assumptions: Total chemists available = 10. Total politicians available = 5. Total economists available = 8. Total biologists available = 15. Team size required = 6 people. At least 5 chemists must be on the team. All individuals are distinct and each person can be selected at most once. Concept / Approach: The phrase “at least 5 chemists” means the team can have either exactly 5 chemists or exactly 6 chemists. We handle each possible case separately and then add the counts, because the cases are mutually exclusive but collectively exhaustive. Within each case, the members are chosen using combinations since the order in which people are picked does not matter; the team is an unordered group. The other non chemist members come from the remaining categories combined as a single pool of distinct people. Step-by-Step Solution: Step 1: Total non chemist pool size = politicians + economists + biologists = 5 + 8 + 15 = 28. Case 1: Exactly 5 chemists and 1 non chemist. Choose 5 chemists from 10: C(10, 5). Choose 1 non chemist from 28: C(28, 1). Number of teams in Case 1 = C(10, 5) * C(28, 1). C(10, 5) = 252 and C(28, 1) = 28, so Case 1 count = 252 * 28 = 7056. Case 2: Exactly 6 chemists and 0 non chemists. Choose 6 chemists from 10: C(10, 6). C(10, 6) = 210. Total number of valid teams = Case 1 + Case 2 = 7056 + 210 = 7266. Verification / Alternative check: Because the team size is 6 and we need at least 5 chemists, there are only two possibilities: (5 chemists, 1 non chemist) or (6 chemists, 0 non chemists). There is no other composition that satisfies the condition. The combination values C(10, 5) and C(10, 6) can be checked manually using factorial definitions. Adding these two non overlapping cases gives 7266, which is consistent with combinatorial reasoning and the problem statement. Why Other Options Are Wrong: Values like 6379 or 6400 do not correspond to any clean combination expression and suggest arithmetic or counting mistakes. The value 7350 is larger than the correct answer and might come from including teams with fewer than 5 chemists. The remaining distractor 7000 is also not equal to C(10, 5) * 28 plus C(10, 6). None of these options correctly model both allowed cases and therefore they are incorrect. Common Pitfalls: Learners often misinterpret “at least 5 chemists” as “exactly 5 chemists” and forget the case with 6 chemists. Another common error is to separate the non chemists by category and overcomplicate the selection rather than treating all non chemists as one pool. Some also mistakenly use permutations instead of combinations even though the team is unordered. Clearly writing out the cases and using combinations for each avoids these mistakes. Final Answer: The number of different 6 person research teams that can be formed with at least 5 chemists is 7266.

Question 10

There are 7 men and 10 women in a selection pool for a committee. The committee will have three distinct positions President, Vice President and Treasurer. In how many different ways can the committee be formed if exactly two of the three positions …

Accepted Answer

Introduction / Context: This problem combines combinations and permutations when forming a small committee with designated positions. Unlike a team where order does not matter, here the roles of President, Vice President and Treasurer are distinct, so different assignments of the same people to different positions count as different committees. The condition that exactly two of these three positions must be filled by men forces us to consider the gender composition and the role assignments carefully. Given Data / Assumptions: Total men available = 7. Total women available = 10. Three distinct positions: President, Vice President, Treasurer. Exactly two positions must be held by men and one by a woman. All individuals are distinct, and one person can hold only one position. Concept / Approach: We must first choose which two men and which one woman will serve, and then assign these three chosen people to the specific positions. Choosing the people is a combinations step because order is not yet relevant. Assigning them to named positions is a permutations step because each different assignment to the roles produces a new arrangement. The total number of valid committees is the product of these two stages. Step-by-Step Solution: Step 1: Choose 2 men out of 7. Number of ways = C(7, 2) = 21. Step 2: Choose 1 woman out of 10. Number of ways = C(10, 1) = 10. Step 3: Now there are 3 chosen people (2 men and 1 woman). Assign them to the three distinct posts. Number of ways to assign 3 distinct people to 3 positions = 3! = 6. Step 4: Multiply all stages together. Total number of valid committees = C(7, 2) * C(10, 1) * 3!. Total = 21 * 10 * 6 = 1260. Verification / Alternative check: An alternative viewpoint is to first select a woman for one of the three posts and then fill the remaining two posts with men. However, because the three posts are symmetric at the beginning, the method above already accounts for all possibilities with a clean combination followed by a permutation. Checking each step numerically confirms that 21 * 10 = 210 ways to choose the people, and for each such choice there are 6 different assignments to positions, giving 210 * 6 = 1260 arrangements. Why Other Options Are Wrong: The values 1200, 1240, 1620 and 840 do not match the correct combination and permutation breakdown. For example, 840 might arise if the assignment to roles is undercounted, while 1620 might come from miscounting the number of ways to choose the men or women. None of these values equals C(7, 2) * C(10, 1) * 3!, so they are incorrect. Common Pitfalls: Students often forget to account for the fact that the three positions are distinct and mistakenly use only combinations for all steps. Others may accidentally choose 2 men and 1 woman but then assign only some of them to roles, ignoring permutations. Another error is accidentally allowing 3 men or 3 women, which violates the “exactly two men” condition. Explicitly separating the selection of people from the assignment of posts prevents these mistakes. Final Answer: The committee with exactly two men and one woman in the three distinct posts can be formed in 1260 different ways.

Question 11

In a league there are 8 teams and each team must play every other team exactly three times during the season. How many matches will be played in total in the league?

Accepted Answer

Introduction / Context: This question is a standard application of combinations in counting pairwise matches in a league. When teams in a round robin tournament all play one another, each match corresponds to an unordered pair of distinct teams. The twist here is that each pair of teams plays not once but three times. The goal is to determine the total number of matches played across the entire league. Given Data / Assumptions: Total number of teams in the league = 8. Each pair of distinct teams plays exactly three matches against each other. There are no extra matches beyond these scheduled ones. Matches are counted without regard to home or away status; each individual game is one match between two teams. Concept / Approach: First we count how many different pairs of teams can be formed from 8 teams. This is a combinations problem because a match between Team A and Team B is the same as a match between Team B and Team A. Once we know the number of unique pairs, we multiply by 3 because each pair plays three matches. This is a straightforward use of the combination formula C(n, 2) for counting unordered pairs. Step-by-Step Solution: Step 1: Compute the number of distinct pairs of teams. Number of ways to choose 2 teams from 8 = C(8, 2). C(8, 2) = 8 * 7 / 2 = 28. Step 2: Each pair of teams plays exactly 3 matches. So, total matches = number of pairs * matches per pair. Total matches = 28 * 3 = 84. Verification / Alternative check: You can also view this as first thinking about a single round robin where each pair meets once, giving 28 matches. The question states that they play each other three times, which is equivalent to three such round robins. Thus, the total number of matches is 3 times 28, again confirming 84 as the correct count. Why Other Options Are Wrong: 36 is too small and corresponds roughly to counting only some of the pairs or using the wrong formula. The number 72 would be correct if each pair played only about 2.57 matches, which is not meaningful here. The value 79 is arbitrary and does not arise from any simple combination calculation. Similarly, 96 would correspond to 32 pairs, but there are only 28 distinct pairs possible among 8 teams. Common Pitfalls: A typical mistake is to confuse ordered and unordered selections and incorrectly use permutations instead of combinations. Another common error is to forget to multiply by 3 after counting the pairs once. Some students may also try to count matches per team individually and double count, which leads to an incorrect total. Remembering that each match is associated with a unique unordered pair and then scaling by the number of games per pair avoids these problems. Final Answer: The league will have a total of 84 matches played.

Question 12

In how many different ways can five distinct persons stand in a single straight line for a group photograph?

Accepted Answer

Introduction / Context: This problem is a direct application of the idea of permutations. We are asked to count the number of ways that five distinct people can be arranged in a straight line. Since each arrangement corresponds to a unique ordering of the five individuals, the question becomes a simple permutation of distinct objects taken all at once. Given Data / Assumptions: There are 5 different persons. They are to stand in a single straight line. Every person is distinct so changing their positions creates a different arrangement. All 5 positions in the line must be filled. Concept / Approach: When arranging n distinct objects in a sequence, the total number of permutations is n factorial (written as n!). This is because there are n choices for the first position, n minus one choices for the second position, and so on down to 1 choice for the last position. For this problem, n equals 5. Hence, the result is 5! which counts all possible standing orders of the five people. Step-by-Step Solution: Step 1: Determine the number of options for the first position. There are 5 people, so there are 5 choices for the first spot. Step 2: Determine the options for the second position. After one person is placed, 4 people remain, so there are 4 choices for the second spot. Step 3: Continue this reasoning. Third spot: 3 choices; fourth spot: 2 choices; fifth spot: 1 choice. Step 4: Multiply the number of choices for each position. Total arrangements = 5 * 4 * 3 * 2 * 1 = 5! = 120. Verification / Alternative check: You can also express the count directly as the permutation formula P(5, 5) which equals 5!, because we are arranging all 5 people. Calculating 5! explicitly confirms 120 distinct arrangements. There is no constraint on who can stand where, so no further adjustments are needed. Why Other Options Are Wrong: The values 240, 360 and 720 are all multiples of 120 that would arise only if extra factors were mistakenly included, such as incorrectly counting additional positions. The value 60 is half of 120 and might come from stopping the multiplication too early or ignoring some positions. Only 120 corresponds exactly to 5!. Common Pitfalls: Some learners misinterpret the question and try to apply combinations, which count selections without order, rather than permutations, which count arrangements with order. Others may incorrectly think that rotating the whole line does not produce new arrangements, but in a line every distinct order counts separately. Remember that whenever order matters for distinct entities, factorial based permutations are usually the correct tool. Final Answer: Five distinct persons can stand in a straight line in 120 different ways.

Question 13

First, second and third prizes are to be awarded at an engineering fair in which 13 different exhibits have been entered. In how many different ways can these three distinct prizes be awarded to the exhibits?

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Introduction / Context: This question is about assigning distinct prizes to different exhibits. Because first, second and third prizes are different in prestige, the order in which exhibits receive these prizes matters. Therefore we are dealing with permutations, not combinations. The set of exhibits is fixed and has 13 distinct entries, and we must determine how many ways we can pick and order three of them to receive the prizes. Given Data / Assumptions: Total exhibits entered in the fair = 13. There are three distinct prizes: first, second and third. Each exhibit can receive at most one prize. The same exhibit cannot receive more than one position. Concept / Approach: We are choosing and ordering 3 exhibits from a set of 13 distinct exhibits. The number of such ordered selections is given by permutations, written as P(13, 3). This can be calculated as 13 * 12 * 11, representing the number of ways to assign the three positions sequentially from the pool of exhibits. Step-by-Step Solution: Step 1: Choose an exhibit for the first prize. There are 13 possible exhibits that can receive first prize. Step 2: Choose an exhibit for the second prize. After the first prize is awarded, only 12 exhibits remain. Step 3: Choose an exhibit for the third prize. Now 11 exhibits remain eligible for third prize. Step 4: Multiply the number of choices for each prize. Total number of prize assignments = 13 * 12 * 11. Compute: 13 * 12 = 156; then 156 * 11 = 1716. Verification / Alternative check: Using the permutation formula, we have P(13, 3) = 13! / (13 - 3)! = 13! / 10!. This equals 13 * 12 * 11, which matches the step by step multiplication above. Both approaches give 1716, confirming the correctness of the result. Why Other Options Are Wrong: The value 1736 differs from 1716 by 20 and does not correspond to any standard permutation calculation. The values 1216 and 1346 are also arbitrary and are not equal to 13 * 12 * 11. The option 2197 equals 13^3, which would be relevant if we allowed repetition of exhibits across prizes, but that is not allowed here because each exhibit can receive at most one prize. Common Pitfalls: A common mistake is to use combinations instead of permutations, calculating C(13, 3) instead of P(13, 3). Combinations ignore the order of selection, whereas here the order (first, second, third) is crucial. Another pitfall is to miscalculate the product 13 * 12 * 11 or to allow an exhibit to win multiple prizes. Keeping the distinction between ordered and unordered selections clear avoids these errors. Final Answer: The three distinct prizes can be awarded in 1716 different ways.

Question 14

When John arrives in New York, he has 8 different tourist stops that he could visit, but he has time to visit only 6 of them. If the order in which he visits the 6 stops matters (it is a schedule), then in how many different ways can he choose and a…

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Introduction / Context: This problem asks us to count how many different itineraries John can follow if he chooses and orders a subset of tourist stops. Because both the choice of which locations he visits and the order in which he visits them matter, we are dealing with permutations of a subset of the available stops. This kind of question appears frequently in arrangement and scheduling contexts in aptitude tests. Given Data / Assumptions: Total distinct tourist stops available = 8. John will visit exactly 6 stops. The order of visiting the stops is important. No stop is visited more than once. Concept / Approach: We need the number of permutations of 8 distinct items taken 6 at a time. This is denoted by P(8, 6) and equals 8! divided by (8 - 6)!, that is 8! / 2!. Conceptually, John first chooses 6 stops from the 8, and then arranges these 6 in all possible orders, so the result captures both selection and ordering. Step-by-Step Solution: Step 1: Use the permutation formula for selecting and ordering 6 stops out of 8. P(8, 6) = 8! / (8 - 6)! = 8! / 2!. Step 2: Compute 8! = 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 = 40320. Step 3: Compute 2! = 2. Step 4: Divide to obtain the number of permutations. P(8, 6) = 40320 / 2 = 20160. Verification / Alternative check: Alternatively, you can reason position by position. For the first stop, John can choose any of 8 places. For the second stop, 7 choices remain, then 6 for the third, and so on down to 3 choices for the sixth stop. The product is 8 * 7 * 6 * 5 * 4 * 3, which equals 20160. This matches the permutation approach and confirms the final answer. Why Other Options Are Wrong: Values such as 20610, 21000 or 24000 are close to 20160 but not equal to the exact product 8 * 7 * 6 * 5 * 4 * 3. They indicate arithmetic mistakes or using extra or fewer factors. The value 13440 would arise if only five positions were considered or if some choices were miscounted. None of these incorrect options equals P(8, 6). Common Pitfalls: One frequent mistake is to treat the problem as a combination and use C(8, 6), which counts only which stops are visited, not their order. Another error is to truncate the product too early or incorrectly include extra multiplication factors. By clearly recognising that the question involves both selection and ordering of stops, and by writing out the full product, we avoid these pitfalls. Final Answer: John can choose and arrange his 6 stop schedule in 20160 different ways.

Question 15

A local delivery company has three packages to deliver to three different houses. Each house is supposed to receive exactly one specific package. If the driver delivers the three packages at random, in how many possible ways will at least one house …

Accepted Answer

Introduction / Context: This problem is about counting permutations with a condition involving correct and incorrect placements. There are three distinct packages and three distinct houses. Each package has a unique intended destination. We are asked to count the number of ways to assign the packages to the houses so that at least one house gets the wrong package. This is an example of using the complement principle in counting, where it can be easier to count the opposite event and subtract from the total. Given Data / Assumptions: There are 3 distinct packages. There are 3 distinct houses. Each package must be delivered to exactly one house. Every arrangement corresponds to a permutation of the 3 packages among the 3 houses. We want the number of permutations where at least one package is misdelivered. Concept / Approach: The total number of possible delivery patterns is the number of permutations of 3 distinct objects, which is 3! = 6. Among these, there is exactly 1 arrangement in which every package reaches its correct house. All the other arrangements have at least one package delivered to the wrong house. Instead of counting all the incorrect patterns directly, we use the complement principle: subtract the number of completely correct arrangements from the total number of arrangements. Step-by-Step Solution: Step 1: Compute the total number of ways to assign the 3 packages to 3 houses. Total permutations = 3! = 3 * 2 * 1 = 6. Step 2: Count the number of perfectly correct deliveries. There is exactly 1 way in which each package goes to its intended house. Step 3: Use the complement principle. Number of ways with at least one wrong delivery = total ways minus all correct ways. So, required count = 6 - 1 = 5. Verification / Alternative check: We can list all permutations of three packages to verify this result. There is 1 arrangement with no mistakes and 2 arrangements where exactly one package is correct and the other two are swapped, plus 3 arrangements where no package is in its correct place. This gives 1 + 2 + 3 = 6 total, with 5 of them having at least one wrong package. This enumeration matches the complement calculation. Why Other Options Are Wrong: The value 6 represents the total number of permutations and would be correct only if every arrangement had at least one wrong package, which is not true because there is one perfect arrangement. The values 3 and 2 are too small and would correspond to miscounting specific partial cases. The value 120 is 5!, which is unrelated to this problem since we have only 3 packages and 3 houses. Common Pitfalls: A frequent mistake is to forget the existence of the fully correct arrangement and simply assume that all permutations correspond to at least one error. Another mistake is to confuse this with the count of permutations where no package is correct, which is called a derangement. In this small example, there are 2 derangements, but the question asks for at least one wrong package, which includes both partial and complete misdelivery patterns. The complement method keeps the logic clear and avoids such confusion. Final Answer: There are 5 different ways in which at least one house receives the wrong package.

Question 16

A group consists of four couples, that is, four men each with his own wife. In how many different ways can they be arranged in a straight line such that men and women occupy alternate positions?

Accepted Answer

Introduction / Context: This arrangement problem involves both gender and positional constraints. We have four men and four women, each man married to one of the women. They must be lined up in such a way that men and women alternate. The key challenge is to count the number of valid orderings that satisfy the alternating pattern, taking into account that the starting position can be either a man or a woman. Given Data / Assumptions: Total men = 4. Total women = 4. Each man has exactly one wife, but the constraint is only about gender, not keeping couples together. They are to stand in a straight line of 8 positions. Men and women must alternate in the line. Concept / Approach: An alternating line of 4 men and 4 women can start with either a man or a woman. So there are two possible gender patterns: M W M W M W M W or W M W M W M W M. For each pattern, we first arrange the men in the positions marked for men and then arrange the women in the positions marked for women. Because all individuals are distinct, we use permutations for each gender group and then multiply, finally accounting for the two possible starting patterns. Step-by-Step Solution: Step 1: Count arrangements if the line starts with a man. For pattern M W M W M W M W, there are 4 positions for men and 4 positions for women. Number of ways to arrange the 4 men = 4! = 24. Number of ways to arrange the 4 women = 4! = 24. Total arrangements for this pattern = 4! * 4! = 24 * 24 = 576. Step 2: Count arrangements if the line starts with a woman. The pattern W M W M W M W M also has 4 positions for women and 4 positions for men. Arrangements again = 4! * 4! = 576. Step 3: Add the contributions of both patterns. Total valid arrangements = 576 + 576 = 1152. Verification / Alternative check: We can factor the calculation as follows: there are 2 choices for whether the first position is occupied by a man or a woman. Once that is fixed, the 4 positions of a given gender can be filled in 4! ways and the other gender positions can be filled in another 4! ways. Hence the total equals 2 * 4! * 4! = 2 * 24 * 24 = 1152. This is a concise confirmation of the result. Why Other Options Are Wrong: The values 1278 and 1296 do not correspond to any simple product of factorials under the alternating constraint. The value 432 is much smaller and would result if one of the 4! factors or the factor of 2 was omitted. The option labeled “None of these” is incorrect because we have found a valid count of 1152 that matches one of the choices. Common Pitfalls: Students sometimes ignore the possibility of starting with a woman and count only one alternating pattern, leading to exactly half the correct answer. Others mistakenly treat the problem as a simple permutation of 8 distinct people without enforcing the alternating gender pattern, which gives 8! and is far too large. Being careful to first fix the gender pattern, then permute within each group, ensures a correct solution. Final Answer: They can be arranged in a straight line with men and women alternating in 1152 different ways.

Question 17

When all the integers from 1 to 1000 (inclusive) are written down, how many times does the digit 8 appear in total?

Accepted Answer

Introduction / Context: Digit counting problems are common in aptitude tests and help develop careful combinatorial reasoning. Here we want to know how many times the digit 8 appears when every integer from 1 to 1000 is written. Each integer can contribute zero, one or more occurrences of the digit 8, and we must count all of them across the entire range. Given Data / Assumptions: We consider all integers from 1 to 1000 inclusive. The numbers are written in standard decimal notation. We count each occurrence of the digit 8 in any position (units, tens, hundreds). The number 1000 contains no digit 8. Concept / Approach: It is easier to treat numbers from 000 to 999 and count occurrences of 8 in each digit place, then adjust for the inclusion of 1000. For 000 to 999, there are 1000 numbers with three positions (hundreds, tens, units). In each position, each digit from 0 to 9 appears equally often. Once we know how many eights appear in each position, we sum across positions. Finally we note that 1000 does not add any eights, so the count remains unchanged for the range 1 to 1000. Step-by-Step Solution: Step 1: Consider all 3 digit strings from 000 to 999. There are 1000 such numbers, and each position (hundreds, tens, units) cycles through the digits 0 to 9 equally. Step 2: Count the number of times digit 8 appears in a fixed position. In any fixed position, each digit occurs in exactly 1 out of every 10 numbers. Therefore, in 1000 numbers, 8 appears 100 times in each position. Step 3: There are 3 positions (hundreds, tens, units). Total occurrences of 8 from 000 to 999 = 100 (hundreds) + 100 (tens) + 100 (units) = 300. Step 4: Adjust from 000 to 999 to 1 to 1000. The leading 0 in 000 does not change the count of digit 8 compared with starting at 1. The number 1000 has no digit 8, so the total remains 300. Verification / Alternative check: To confirm, we could test a smaller range, such as 1 to 100, and verify that the same positional argument works there. For 1 to 100, each digit appears 20 times in each position among the numbers 00 to 99, leading to 20 eights per position and 60 in total. Extending this reasoning to 000 to 999 is logically consistent. The absence of any 8 in 1000 makes the 300 count final. Why Other Options Are Wrong: The options 100 and 200 underestimate the count because they consider only one or two positions or miscount the frequency of 8. The value 400 would overcount, as it implies more than 100 eights per position in 1000 numbers, which is impossible. The option 180 is similarly too small and does not align with equally distributed digits. Common Pitfalls: A typical mistake is to attempt to manually count ranges like 80 to 89 or 800 to 899 and double count overlaps such as 888. Others might forget that numbers less than 100 or 10 can be treated with leading zeros to apply the positional method. Another common error is to ignore that each digit appears equally often in each place when all numbers are considered systematically. Using the positional counting method avoids such errors. Final Answer: The digit 8 appears a total of 300 times when the integers from 1 to 1000 are listed.

Question 18

There are 41 students in a class, and the number of girls is one more than the number of boys. A team of four students is to be formed such that all four team members are not of the same gender and the numbers of girls and boys in the team are not e…

Accepted Answer

Introduction / Context: This problem combines combinatorics with constraints based on gender composition. We must form a four student team from a class where the counts of boys and girls are closely related. However, the team is restricted: it cannot be all boys, it cannot be all girls, and it also cannot have an equal number of boys and girls. This forces us to carefully enumerate the allowed gender combinations and count the selections for each case. Given Data / Assumptions: Total students = 41. Number of girls is one more than the number of boys. Let number of boys be B and number of girls be G. B + G = 41 and G = B + 1. Team size = 4 students. Disallowed teams: all four from same gender, and teams with equal numbers of boys and girls. Concept / Approach: First, we solve for the numbers of boys and girls. Then we list all possible gender splits for a team of four and remove those that violate the conditions. The valid splits are those where there is at least one of each gender and the counts are not equal. After identifying valid splits, we compute the number of teams for each split using combinations and then sum the results. Step-by-Step Solution: Step 1: Solve for the number of boys and girls. B + G = 41 and G = B + 1. Substitute G: B + (B + 1) = 41, so 2B + 1 = 41, giving 2B = 40 and B = 20. Thus, G = B + 1 = 21. Step 2: List all possible gender splits for a team of 4. Possible splits: 4 boys 0 girls, 3 boys 1 girl, 2 boys 2 girls, 1 boy 3 girls, 0 boys 4 girls. Conditions: not all same gender and not equal numbers of boys and girls. Therefore, allowed splits are 3 boys 1 girl and 1 boy 3 girls. Step 3: Count teams with 3 boys and 1 girl. Choices for boys: C(20, 3). Choices for girls: C(21, 1). Number of teams = C(20, 3) * C(21, 1) = 1140 * 21 = 23940. Step 4: Count teams with 1 boy and 3 girls. Choices for boys: C(20, 1) = 20. Choices for girls: C(21, 3). C(21, 3) = 1330. Number of teams = 20 * 1330 = 26600. Step 5: Add both valid cases. Total valid teams = 23940 + 26600 = 50540. Verification / Alternative check: We can quickly check the disallowed cases. Teams of 4 boys: C(20, 4). Teams of 4 girls: C(21, 4). Teams of 2 boys and 2 girls: C(20, 2) * C(21, 2). None of these should be included. Counting only 3 boys 1 girl and 1 boy 3 girls, as we have done, matches the logical constraints. The final total 50540 is consistent with the combinatorial calculations. Why Other Options Are Wrong: The values 49450, 46587 and 52487 do not equal the sum of the two allowed case counts and likely include some disallowed configurations or arithmetic errors. The neat number 48000 is also incorrect and does not match the precise combination calculations for the valid gender splits. Common Pitfalls: Many learners forget to exclude teams with equal numbers of boys and girls or accidentally allow all boy or all girl teams. Others miscompute C(21, 3) or C(20, 3), leading to small errors that change the final total. Systematically listing all possible splits, marking which are allowed and which are disallowed, and then applying the combination formula carefully helps avoid such mistakes. Final Answer: The team of four students can be formed in 50540 different ways under the given gender constraints.

Question 19

How many distinct 4 digit even numbers can be formed using the digits 2, 7, 5, 3, 9 and 1 if no digit is repeated in a number?

Accepted Answer

Introduction / Context: This question tests understanding of how to form even numbers under digit and repetition constraints. We have a fixed set of digits and must form 4 digit numbers that are even and do not repeat digits. Even numbers are characterised by having an even digit in the units place, so we start by deciding which digits can appear at the end and then count valid permutations. Given Data / Assumptions: Available digits: 2, 7, 5, 3, 9, 1. We must form 4 digit numbers. No digit may repeat within a number. A valid number must be even. All numbers are in standard decimal notation with no leading zero (and here there is no zero provided). Concept / Approach: A number is even if its last digit is even. Among the given digits, only 2 is even. Therefore, for any 4 digit even number using these digits exactly once, the units digit must be 2. Once the units position is fixed, we choose and arrange the remaining three positions (thousands, hundreds and tens) using the remaining five digits, with no repetition. This is a permutation problem for the remaining positions. Step-by-Step Solution: Step 1: Identify even digits in the given set. Digits: 2, 7, 5, 3, 9, 1. Only 2 is even. Step 2: Fix the units position. To make the number even, the units digit must be 2. Step 3: Determine the remaining digits and positions. Remaining digits to fill the first three positions: 7, 5, 3, 9, 1 (5 digits). Positions to be filled: thousands, hundreds, tens (3 positions). Step 4: Count the permutations of these 5 digits taken 3 at a time. Number of ways = P(5, 3) = 5 * 4 * 3 = 60. Each such arrangement, followed by the fixed trailing digit 2, gives a unique 4 digit even number. Verification / Alternative check: We can confirm the logic by noting that there are 5 choices for the first digit, 4 choices for the second digit and 3 choices for the third digit, since we cannot reuse digits. After choosing these three, the last digit 2 is fixed. Multiplying 5 * 4 * 3 equals 60, which matches the permutation calculation and confirms the result. Why Other Options Are Wrong: Values such as 59, 61 and 64 do not match the clean product 5 * 4 * 3. They may arise from small arithmetic errors or an incorrect assumption that there are more or fewer valid last digits. The value 48 would result from incorrectly treating the problem as permutations of only 4 digits from 5 without accounting properly for the fixed units digit or omitting some choices. Common Pitfalls: A common mistake is to assume that more than one digit in the set is even or to forget that only the last digit determines evenness. Another frequent error is to treat the first digit differently for fear of a leading zero, but zero is not present in the digits, so there is no restriction beyond non repetition. Correctly fixing the last digit as 2 and then permuting the remaining digits avoids these pitfalls. Final Answer: There are 60 distinct 4 digit even numbers that can be formed under the given conditions.

Question 20

Consider the word RECREATIONAL. If you take the first, seventh, ninth and tenth letters of this word and try to form a meaningful English word using each of those four letters exactly once, what will be the third letter of such a word? (If more than…

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Introduction / Context: This question checks verbal reasoning and attention to letter positions in a word. We must carefully pick specific letters from the given word RECREATIONAL based on their positions, then determine whether they can form a meaningful English word. If a valid word exists, we have to identify the third letter of that word. The question also allows for the possibilities of multiple words or no word at all, represented by special choices X and Z respectively. Given Data / Assumptions: The word is RECREATIONAL. We need letters at positions 1, 7, 9 and 10. We must use each selected letter exactly once. If a meaningful word exists, we report its third letter. If more than one meaningful word exists, we answer X. If no meaningful word exists, we answer Z. Concept / Approach: The first step is positional: identify which letters occur at the specified positions of the word. The second step is to attempt to form meaningful English words from these four letters. By examining common patterns and trying different orders, we can see whether a real word is possible. We then decide if there is a unique meaningful word or multiple possibilities. Finally, we read off the third letter of the word we have identified. Step-by-Step Solution: Step 1: Write the letters of RECREATIONAL with their positions. R(1), E(2), C(3), R(4), E(5), A(6), T(7), I(8), O(9), N(10), A(11), L(12). Step 2: Extract the required letters. Position 1: R. Position 7: T. Position 9: O. Position 10: N. So, the letters are R, T, O and N. Step 3: Try to form a meaningful word using these four letters exactly once. One obvious word is TORN, which is a valid English word. Check other orders: RNOT, TRON, NORT and so on do not form standard English words. TORN appears to be the only common meaningful word. Step 4: Identify the third letter of TORN. The word TORN has letters T (1st), O (2nd), R (3rd), N (4th). Therefore, the third letter is R. Verification / Alternative check: We can systematically list all permutations of the letters R, T, O and N and check which, if any, are standard words: TORN is valid, while forms like TORN, NORT, RTON and others do not appear as normal dictionary words. Because we find only TORN as a familiar English word, we can safely assume there is a unique meaningful word. Therefore, we do not need to choose X or Z. Why Other Options Are Wrong: Option T or N would be correct only if the third letter of the meaningful word was T or N, which it is not. Option X would apply if more than one meaningful word could be made from the letters, but only TORN is standard. Option Z would apply if there were no meaningful word, which is not the case here. The only correct third letter is R, so option R is correct. Common Pitfalls: Some students misread letter positions and pick the wrong characters from the word RECREATIONAL. Others might overlook TORN as a word or mistakenly consider non standard abbreviations as valid words, leading them to choose X. Carefully listing positions, extracting the correct letters and thinking of common four letter words is the key to solving this reliably. Final Answer: The meaningful word formed is TORN, whose third letter is R.

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