Question 1

Number of children visiting an ice cream shop when 9 flavours are available and each child takes a cone with a unique pair of flavours

Accepted Answer

Introduction / Context:This question uses combinations to model how many distinct two flavour cones can be made from a set of flavours. Each child chooses a cone with two distinct flavours, and no two children choose the same pair, so the number of children equals the number of unique two flavour combinations. Given Data / Assumptions: There are 9 distinct ice cream flavours available. Each child chooses exactly one cone. Each cone contains exactly two different flavours. No two children choose the same combination of flavours. Order of flavours on the cone does not matter, so flavour A with flavour B is the same as flavour B with flavour A. Concept / Approach:Each cone is identified by a combination of 2 distinct flavours chosen from 9. Since order does not matter, the correct count is given by a combination, not a permutation. The number of children is equal to the number of such distinct combinations. Step-by-Step Solution:Step 1: Let the number of flavours be n = 9.Step 2: Each cone picks 2 distinct flavours, so the number of possibilities is 9C2.Step 3: Use the combination formula nC2 = n * (n - 1) / 2.Step 4: Substitute n = 9 to get 9C2 = 9 * 8 / 2.Step 5: Compute 9 * 8 = 72 and divide by 2 to get 36.Step 6: No two children can have the same combination, so the maximum number of children is 36. Verification / Alternative check:You can list pairs in groups to verify the count. For example, pair the first flavour with the remaining 8, the second with the next 7, and so on. The total pairs become 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 36, which matches 9C2. Why Other Options Are Wrong: 28, 44, and 56 are other combination values from different nC2 cases but do not match 9C2. 56 would be the number of pairs from 8 distinct flavours plus some incorrect extra counting. Common Pitfalls:A common error is to use permutations 9P2 = 72, which would treat flavour order as important and overcount each cone by a factor of 2. Another mistake is to misread the question and include combinations with repetition, such as two scoops of the same flavour, which is not allowed here. Final Answer:The number of children who can each have a unique two flavour cone is 36, so the correct answer is 36.

Question 2

Number of three digit numbers where exactly two of the digits are the same and the third digit is different

Accepted Answer

Introduction / Context:This problem tests your ability to count three digit numbers under a digit pattern restriction. You must count numbers where exactly two digits are equal and the third digit is different, while still ensuring the number is a valid three digit number. Given Data / Assumptions: We consider only three digit numbers from 100 to 999. Digits are taken from 0 to 9, but the first digit cannot be zero. The pattern required is that exactly two digits are the same and the remaining digit is different. All three digits equal, such as 111, are not allowed. Concept / Approach:We must count numbers with exactly one repeated digit. So we choose the digit that repeats, choose the digit that is different, and then decide which two positions get the repeated digit. Each case must respect the condition that the first digit cannot be zero. Step-by-Step Solution:Step 1: Count cases by position pattern: x x y, x y x, and y x x where x is the repeated digit and y is the different digit.Step 2: For pattern x x y, the first digit is x and must be non zero. So x can be 1 to 9 giving 9 choices.Step 3: For each x, y can be any digit from 0 to 9 except x, giving 9 choices.Step 4: So pattern x x y contributes 9 * 9 = 81 numbers.Step 5: For pattern x y x, first digit is x and must again be non zero, so there are 9 choices for x, and 9 choices for y not equal to x, giving another 81 numbers.Step 6: For pattern y x x, the repeated digit is x in positions two and three, and the first digit y must be non zero and different from x.Step 7: If x is zero, then y can be any of 1 to 9, giving 9 numbers.Step 8: If x is from 1 to 9, there are 9 choices for x and then 8 possible non zero digits for y that are different from x, giving 9 * 8 = 72 numbers.Step 9: So pattern y x x contributes 9 + 72 = 81 numbers.Step 10: Total numbers with exactly two equal digits = 81 + 81 + 81 = 243. Verification / Alternative check:You can also programmatically generate all three digit numbers and count how many fit the pattern, which will again yield 243. This confirms that the manual counting is consistent. Why Other Options Are Wrong: 252 and 648 are incorrect counts that may come from double counting or from allowing all three digits to be equal. 900 is the total number of three digit numbers and ignores any restriction. Common Pitfalls:Some learners accidentally include numbers where all three digits are the same. Others forget the leading digit restriction and treat the first digit as if it could be zero. It is crucial to treat patterns by position and carefully enforce the restriction on the first digit. Final Answer:The number of three digit numbers with exactly two equal digits is 243, so the correct answer is 243.

Question 3

Number of ways to seat 6 boys and 6 girls in a row so that no two girls sit together

Accepted Answer

Introduction / Context:This question tests advanced permutation reasoning with spacing restrictions. You must seat 6 boys and 6 girls in a single row such that no two girls sit next to each other. Problems like this are common in competitive exams and require a clear strategy. Given Data / Assumptions: There are 6 distinct boys. There are 6 distinct girls. All 12 children must be seated in one row. No two girls are allowed to occupy adjacent seats. All boys and girls are distinguishable from one another. Concept / Approach:The standard approach is to first seat the boys, then place the girls in the gaps between or at the ends of boys. Once the boys are placed, they create fixed slots where girls can be inserted. We then choose positions for girls in these slots and arrange the girls. Step-by-Step Solution:Step 1: Arrange the 6 boys in a row in 6! ways.Step 2: After seating boys, there are 7 possible slots where girls can sit: one before the first boy, one after the last boy, and one between every pair of consecutive boys.Step 3: To ensure that no two girls sit together, we can place at most one girl in each slot.Step 4: We have 6 girls and 7 distinct slots, so we must choose 6 different slots from these 7 and then assign the girls to those chosen slots.Step 5: Number of ways to pick ordered slots for the 6 distinct girls is 7P6, that is, permutations of 7 slots taken 6 at a time.Step 6: Therefore, the total number of valid seating arrangements is 6! * 7P6. Verification / Alternative check:You can also compute 7P6 as 7! / 1! = 5040 and note that 6! = 720, so the total number of arrangements equals 720 * 5040 which is 3628800. This is a very large but reasonable number given 12 distinct people with restrictions. Why Other Options Are Wrong: 2(6!) and 6! x 7 do not incorporate the full complexity of distributing 6 distinct girls into 7 slots. None is incorrect because the expression 6! x 7P6 correctly models the situation. Common Pitfalls:Students sometimes mistakenly choose only 6 slots out of 7 using combinations 7C6 and then multiply by 6!, which in fact already equals 7P6 but is sometimes misinterpreted. Others forget the spacing idea and attempt to directly arrange all 12 people with complicated inclusion exclusion. The slot method is far simpler and more reliable. Final Answer:The number of ways to seat the group so that no two girls sit together is 6! times 7P6, so the correct answer is 6! x ⁷P₆.

Question 4

Number of ways to form a 5 member team from 7 boys and 6 girls such that the team includes at least 3 girls

Accepted Answer

Introduction / Context:This selection problem uses combinations with an additional constraint on the composition of the team. The team must contain at least three girls, which means we must consider multiple cases for how many girls are included and then sum those cases. Given Data / Assumptions: Total boys available: 7. Total girls available: 6. Team size: 5 members. The team must have at least 3 girls. All boys and girls are distinct individuals. Concept / Approach:At least three girls means the team can have exactly 3 girls, exactly 4 girls, or exactly 5 girls. For each case, we count the number of ways to select the required number of girls and boys using combinations, then add the results because the cases are mutually exclusive. Step-by-Step Solution:Step 1: Case 1 is exactly 3 girls and 2 boys.Step 2: Select 3 girls from 6 in 6C3 ways and 2 boys from 7 in 7C2 ways.Step 3: Compute 6C3 = 20 and 7C2 = 21, so Case 1 count = 20 * 21 = 420.Step 4: Case 2 is exactly 4 girls and 1 boy.Step 5: Select 4 girls from 6 in 6C4 ways and 1 boy from 7 in 7C1 ways.Step 6: Compute 6C4 = 15 and 7C1 = 7, so Case 2 count = 15 * 7 = 105.Step 7: Case 3 is exactly 5 girls and 0 boys.Step 8: Select 5 girls from 6 in 6C5 ways and 0 boys from 7 in 7C0 ways.Step 9: Compute 6C5 = 6 and 7C0 = 1, so Case 3 count = 6 * 1 = 6.Step 10: Total valid teams = 420 + 105 + 6 = 531. Verification / Alternative check:You can verify by ensuring that none of the cases exceed the available boys or girls and that the total team size is always 5. All three cases satisfy these constraints, and their counts add directly because they do not overlap. Why Other Options Are Wrong: 427 and 651 are close in size but do not match the precise sum of 420, 105, and 6. 714 is too large and would require counting extra invalid cases where there are fewer than 3 girls. Common Pitfalls:Students sometimes include teams with 2 girls by mistake, misreading at least three girls as at most three girls. Another common error is to forget the case with 5 girls and 0 boys. Always write out all valid compositions and check that their sum matches the team size. Final Answer:The number of ways to form such a team is 531, so the correct answer is 531.

Question 5

Number of triangles that can be formed from 12 points on a plane when 8 of them are collinear

Accepted Answer

Introduction / Context:This geometry based combinatorics question asks you to count the number of triangles that can be formed from a set of points, some of which are collinear. Triangles require three non collinear points, so any triple of collinear points cannot form a triangle. Given Data / Assumptions: There are 12 points in total on a plane. Out of these, 8 points are collinear, that is, they lie on the same straight line. The remaining 4 points are not on that line and are in general position so that no additional sets of three points are collinear. A triangle is formed by choosing 3 points that are not collinear. Concept / Approach:First, count all possible triples of points that can be chosen from the 12 points. Then, subtract the triples that do not form triangles, which are precisely the triples taken entirely from the 8 collinear points. This is a standard use of combinations with exclusion of degenerate cases. Step-by-Step Solution:Step 1: Count all possible triples of points from 12 points using 12C3.Step 2: Compute 12C3 = 12 * 11 * 10 / (3 * 2 * 1) = 220.Step 3: Now consider the 8 collinear points. Any three chosen from these form a straight line, not a triangle.Step 4: The number of such collinear triples is 8C3.Step 5: Compute 8C3 = 8 * 7 * 6 / (3 * 2 * 1) = 56.Step 6: Valid triangles are all triples minus collinear triples.Step 7: Number of triangles = 220 - 56 = 164. Verification / Alternative check:You could also think in terms of constructing triangles that include at least one of the 4 non collinear points. It is more complex but should lead to the same total. The subtraction method is much more straightforward and widely used in such problems. Why Other Options Are Wrong: 220 is the total number of triples and ignores the fact that some triples are collinear and do not form triangles. 20 and 160 are far from the correct value and do not match any of the combination steps. Common Pitfalls:A common mistake is to forget that three collinear points do not form a triangle. Another error is to subtract the wrong number of collinear triples, for example using 8C2 instead of 8C3. Always ensure that you use combination counts that match the size of the sets you are excluding. Final Answer:The number of triangles that can be formed is 164, so the correct answer is 164.

Question 6

Number of ways to arrange 4 fairy tale books, 5 novels and 3 plays on a shelf when each category stays grouped and the order of categories is fixed

Accepted Answer

Introduction / Context:This arrangement problem involves books in three categories that must remain grouped, with the groups themselves arranged in a fixed order. We must count how many ways we can reorder books within each group while preserving the category ordering. Given Data / Assumptions: There are 4 distinct books of fairy tales. There are 5 distinct novels. There are 3 distinct plays. On the shelf, all fairy tales must be placed together, followed by all novels together, followed by all plays together, in that fixed category order. Within each category, the order of individual books can vary. Concept / Approach:Since the categories must appear in the fixed order fairy tales, then novels, then plays, there is no rearrangement of groups. The only freedom is in arranging the books within each group. Therefore, the total number of arrangements is the product of the permutations within each group. Step-by-Step Solution:Step 1: Arrange the 4 distinct fairy tale books among themselves.Step 2: The number of ways to arrange 4 distinct items is 4! = 24.Step 3: Arrange the 5 distinct novels among themselves.Step 4: The number of ways to arrange 5 distinct items is 5! = 120.Step 5: Arrange the 3 distinct plays among themselves.Step 6: The number of ways to arrange 3 distinct items is 3! = 6.Step 7: Since the groups must appear in a fixed order, the total number of arrangements is 4! * 5! * 3!.Step 8: Compute 24 * 120 = 2880 and then 2880 * 6 = 17280. Verification / Alternative check:If the category order was not fixed, we would also multiply by 3! for the permutations of the three groups and get 103680. The question specifies the order, so we only consider internal arrangements of each group, confirming that 17280 is correct. Why Other Options Are Wrong: 12400, 17820, and 12460 are similar sized numbers but do not equal 4! * 5! * 3!. 17820 is close but not exact, indicating a likely arithmetic error from some partial computation. Common Pitfalls:Some students assume that the categories can also be rearranged and therefore multiply by 3!, giving a larger number. Others forget that all books are distinct and undercount by using combinations instead of permutations. Always interpret whether group order is fixed or variable. Final Answer:The number of valid shelf arrangements is 17280, so the correct answer is 17280.

Question 7

Minimum number of alphabets required in a language to create at least 1 million distinct three letter initials

Accepted Answer

Introduction / Context:This problem relates to counting the number of possible three letter initials you can form from an alphabet of a given size. It is similar to counting license plates or codes and tests your understanding of basic permutation counting with repetition allowed. Given Data / Assumptions: We want to form three letter initials of the form X Y Z. Each position in the initial can be any letter from the alphabet. Letters can repeat in an initial. We need at least 1 million distinct three letter initials. The alphabet size, say n, is the same for each position. Concept / Approach:When each of the 3 positions can independently be any of n letters, the total number of distinct three letter strings is n^3. The requirement is that n^3 should be at least 1 million. We therefore find the smallest integer n such that n^3 is greater than or equal to 1000000. Step-by-Step Solution:Step 1: Let n be the number of different letters in the alphabet.Step 2: The number of possible three letter initials is n^3 because each of the 3 positions can be filled in n ways.Step 3: We need n^3 greater than or equal to 1000000.Step 4: Observe that 100^3 = 100 * 100 * 100 = 1000000 exactly.Step 5: For any integer n less than 100, such as 99, the cube n^3 is less than 1000000.Step 6: Therefore the smallest integer n that satisfies n^3 greater than or equal to 1000000 is 100. Verification / Alternative check:You can compute 10^3 = 1000 and 50^3 = 125000, which are far below 1 million. Checking 100^3 gives exactly 1 million, confirming that this alphabet size is just enough to reach the requirement. Why Other Options Are Wrong: 1000 would give 1000^3, which is 1000 billion, far more than needed; it is not the minimum. 500 would give 125 million three letter initials, again too large and not minimal. 999 is also unnecessarily large and would exceed the requirement by a huge margin. Common Pitfalls:Some learners misinterpret the requirement and think that the number of initials must be exactly equal to 1 million rather than at least 1 million. Others mistakenly use nC3 instead of n^3, which would be appropriate only if letters could not repeat and order did not matter. Here, order clearly matters and repetition is allowed. Final Answer:The minimum alphabet size required is 100 letters, so the correct answer is 100.

Question 8

In a convex polygon that has 7 sides, how many distinct diagonals can be drawn between its vertices?

Accepted Answer

Introduction / Context:This question tests your understanding of how to count the number of diagonals in a polygon using a standard combinatorial formula. Diagonals are line segments that join two non-adjacent vertices of a polygon, and knowing how to compute them quickly is a common aptitude and competitive exam skill. Given Data / Assumptions: The polygon is convex and has 7 sides (a heptagon). Every pair of non-adjacent vertices can be joined to form exactly one diagonal. Sides of the polygon are not counted as diagonals. Concept / Approach:The total number of straight line segments that can be drawn between n vertices is given by the combination nC2, because each segment is determined by choosing 2 distinct vertices. Among these, n segments are sides of the polygon. The remaining segments are diagonals. There is also a direct formula for diagonals in an n sided polygon: number of diagonals = n * (n - 3) / 2. Step-by-Step Solution:Let n = 7 for a 7 sided polygon.Use the diagonal formula: diagonals = n * (n - 3) / 2.Substitute n = 7: diagonals = 7 * (7 - 3) / 2.Compute the expression: 7 * 4 / 2 = 28 / 2.Therefore, number of diagonals = 14. Verification / Alternative check:You can also compute total segments: 7C2 = (7 * 6) / 2 = 21.Subtract the 7 sides to get diagonals: 21 - 7 = 14.Both methods give the same result, confirming the answer. Why Other Options Are Wrong:7 is just the number of sides, not the number of diagonals.15 is slightly larger than the correct count and could come from careless arithmetic on the formula.21 is the total number of line segments between vertices, including sides, so it overcounts. Common Pitfalls:Students sometimes forget to subtract the sides from the total segments.Another common mistake is using n * (n - 1) instead of nC2.Some may misremember the formula as n * (n - 2) / 2, which is incorrect for diagonals. Final Answer:The number of diagonals in a 7 sided polygon is 14.

Question 9

On a circle there are 15 distinct points; how many different triangles can be formed by joining these points as vertices?

Accepted Answer

Introduction / Context:This problem checks your understanding of combinations and basic counting principles. When you form triangles from points on a circle, you simply need to count how many ways you can select 3 distinct points out of the total points, because any three non-collinear points form exactly one unique triangle. Given Data / Assumptions: There are 15 distinct points on a circle. No three points are collinear except on the circle itself, so any 3 points chosen form a valid triangle. We are counting triangles only by their vertex sets, not by drawing order or orientation. Concept / Approach:To form a triangle, you must choose 3 vertices from the 15 available points. The order in which you choose these 3 vertices does not matter; hence we use combinations, not permutations. The relevant formula is nC3 = n * (n - 1) * (n - 2) / 6 for the number of ways to choose 3 items from n items without regard to order. Step-by-Step Solution:We need to compute 15C3.Use the formula: 15C3 = 15 * 14 * 13 / 6.First compute the numerator: 15 * 14 = 210, and 210 * 13 = 2730.Now divide by 6: 2730 / 6 = 455.Therefore, the number of distinct triangles = 455. Verification / Alternative check:We can simplify during multiplication: 15C3 = (15 * 14 * 13) / (3 * 2 * 1).Cancel 14 with 2 to get 7, and cancel 15 with 3 to get 5.Now compute 5 * 7 * 13 = 35 * 13 = 455.This quick simplification confirms the calculation. Why Other Options Are Wrong:450 is close to the correct value and could come from rounding or subtracting 5 by mistake.469 and 500 do not correspond to any simple combination calculation with 15 points and are clear overestimates. Common Pitfalls:Using permutations (15P3) instead of combinations, which would incorrectly count each triangle multiple times.Wrongly subtracting combinations or assuming some triangles are degenerate, which is not the case for distinct points on a circle.Careless arithmetic when multiplying or dividing large numbers. Final Answer:The number of triangles that can be formed is 455.

Question 10

Six fair coins are tossed at the same time; in how many different outcomes do at most three of the coins show heads?

Accepted Answer

Introduction / Context:This question checks your knowledge of binomial coefficients in the context of coin tossing. When multiple fair coins are tossed simultaneously, each outcome can be described by the number of heads, and counting outcomes with restrictions such as at most three heads is a standard combinatorics task in aptitude exams. Given Data / Assumptions: Six identical fair coins are tossed at the same time. Each coin has two outcomes: Head (H) or Tail (T). We must count the number of outcomes in which the total number of heads is at most three. At most three heads means 0, 1, 2 or 3 heads are allowed. Concept / Approach:The number of outcomes with exactly k heads when n fair coins are tossed is given by the binomial coefficient nCk, because we only need to choose which k of the n coins show heads. To find the outcomes with at most three heads, we sum nCk for k = 0, 1, 2 and 3 when n = 6. Step-by-Step Solution:Total coins n = 6.We want sum of 6C0 + 6C1 + 6C2 + 6C3.Compute: 6C0 = 1 (all tails).6C1 = 6, 6C2 = 15, 6C3 = 20.Add them: 1 + 6 + 15 + 20 = 42.Therefore, there are 42 such outcomes. Verification / Alternative check:Total outcomes for six coins = 2^6 = 64.Outcomes with more than three heads are those with 4, 5 or 6 heads.Compute 6C4 = 15, 6C5 = 6, 6C6 = 1.Sum = 15 + 6 + 1 = 22.Now 64 - 22 = 42, which matches our previous result. Why Other Options Are Wrong:15 is just the value of 6C2 and does not include other allowed cases.16 could result from wrongly assuming half the outcomes qualify.40 is near the correct value but misses two outcomes due to miscounting. Common Pitfalls:Forgetting to include the case with 0 heads.Using permutations instead of combinations when counting head positions.Misreading at most three heads as exactly three heads. Final Answer:The number of outcomes with at most three heads is 42.

Question 11

Eight boxes contain distinct numbers of chocolates from 1 to 8. In how many ways can four boxes be given to four persons, one box to each, so that the first person receives more chocolates than each of the other three, the second person receives mor…

Accepted Answer

Introduction / Context:This problem combines basic combinatorics with ordering constraints. You have boxes with distinct quantities of chocolates and you want to distribute four of them to four labelled persons in such a way that their chocolate counts follow a strict decreasing order from person one to person four. Understanding how to convert such inequality conditions into a simple counting argument is a useful exam skill. Given Data / Assumptions: There are 8 boxes, containing 1, 2, 3, 4, 5, 6, 7 and 8 chocolates respectively. Four persons receive one box each. Let their chocolates be x1, x2, x3 and x4 for persons 1, 2, 3 and 4 respectively. The conditions are: x1 greater than x2, x3 and x4; x2 greater than x3 and x4; and x3 greater than x4. Concept / Approach:The inequalities force a strict ordering: x1 is the largest, x2 is the second largest, x3 is the third largest and x4 is the smallest of the four chosen numbers. Once you choose any set of four distinct box sizes, the assignment to persons is completely fixed by these conditions. Therefore, the count reduces to choosing 4 box sizes out of 8, without regard to order: this is simply 8C4. Step-by-Step Solution:We need to select any 4 distinct numbers out of 8: compute 8C4.Use formula: 8C4 = 8 * 7 * 6 * 5 / (4 * 3 * 2 * 1).Simplify: numerator 8 * 7 * 6 * 5 = 1680; denominator 24.Compute 1680 / 24 = 70.For each such set, the largest goes to person 1, second largest to person 2, third to person 3, and smallest to person 4. Verification / Alternative check:Instead of thinking about assignments, note that any 4 chosen sizes can be uniquely arranged to satisfy the required inequalities.There is exactly 1 valid assignment for each 4 element subset of the 8 box sizes.Thus the final count equals the number of such subsets, which is 8C4 = 70. Why Other Options Are Wrong:40, 72 and 80 do not correspond to any meaningful combination count for selecting 4 of 8 under these constraints.Some of these values may come from misapplying permutations or forgetting that the inequalities already fix the order. Common Pitfalls:Trying to permute the four chosen numbers among the four people even though the inequalities already fix who gets which size.Misinterpreting the condition as only requiring x1 to be maximum and ignoring the full strict chain x1 greater x2 greater x3 greater x4.Computing permutations 8P4, which is much larger and incorrect. Final Answer:The number of valid distributions is 70.

Question 12

If the letters of the word VERMA are written in all possible arrangements in dictionary order, what is the rank (position) of the word VERMA in this ordered list?

Accepted Answer

Introduction / Context:This question examines your understanding of lexicographic (dictionary) order and permutations of distinct letters. The rank of a particular arrangement tells you how many valid permutations come before it when all are listed from smallest to largest in alphabetical order, which is a classic type of reasoning problem in aptitude tests. Given Data / Assumptions: The word is VERMA. The letters are all distinct: V, E, R, M, A. Dictionary order means standard alphabetical order with letters arranged from A to Z. We must count all permutations of these letters that appear before VERMA and then add one to obtain its rank. Concept / Approach:First list the letters in alphabetical order: A, E, M, R, V. Then, position by position, count how many permutations begin with a letter smaller than the corresponding letter of VERMA. For each such smaller starting choice, the remaining letters can permute freely. The total count of these permutations gives the number of words before VERMA. Step-by-Step Solution:Alphabetic order of letters: A, E, M, R, V.First letter V: smaller letters are A, E, M, R (4 letters).For each such choice, the remaining 4 letters can be arranged in 4! = 24 ways. Contribution: 4 * 24 = 96.Now fix V. Remaining letters: A, E, M, R in alphabetical order.Second letter of VERMA is E. Smaller among remaining are A only (1 letter). The remaining 3 letters can arrange in 3! = 6 ways. Add 6 to total: 96 + 6 = 102.Fix V, E. Remaining letters: A, M, R. Third letter is R. Smaller letters are A and M (2 letters). Each yields 2! = 2 arrangements for the last two letters. Contribution: 2 * 2 = 4. Total so far: 106.Fix V, E, R. Remaining letters: A, M. Fourth letter is M. Smaller letter is A (1 letter). With A in fourth position, last letter must be M, so 1 arrangement. New total: 107.The word VERMA itself comes next, so its rank is 107 + 1 = 108. Verification / Alternative check:You can cross check by quickly listing a few final permutations around VERMA to see that exactly 107 distinct permutations precede it.The stepwise positional counting is the standard and reliable method. Why Other Options Are Wrong:117 and 180 are larger ranks that would require more permutations before VERMA than actually exist.810 is far too large, exceeding the total permutations 5! = 120. Common Pitfalls:Forgetting to keep letters in strict alphabetical order at each step when counting smaller possibilities.Not multiplying by factorial of remaining letters after fixing earlier choices.Assuming that the rank is simply the numeric value or position of some letter without doing full combinatorial counting. Final Answer:The rank of the word VERMA is 108.

Question 13

How many different four letter words (not necessarily meaningful) can be formed from the distinct letters of the word LOGARITHMS, if repetition of letters is not allowed?

Accepted Answer

Introduction / Context:This question checks your understanding of permutations of distinct objects. You are given a word with all different letters and asked to count how many four letter sequences you can build without repeating any letter, which is a straightforward application of permutation principles. Given Data / Assumptions: The word LOGARITHMS has 10 distinct letters: L, O, G, A, R, I, T, H, M, S. We must form four letter words using these letters. No letter can be repeated within a word. Order matters because different arrangements such as GLOT and TLOG are considered different words. Concept / Approach:Whenever you are arranging k distinct items selected from n distinct items, and order matters with no repetition, you use permutations. The count is given by nPk = n * (n - 1) * (n - 2) ... up to k factors, or equivalently n! / (n - k)!. Here n = 10 and k = 4. Step-by-Step Solution:Number of distinct letters, n = 10.We need the number of 4 letter arrangements: compute 10P4.Use formula: 10P4 = 10 * 9 * 8 * 7.First multiply 10 * 9 = 90, then 90 * 8 = 720, and 720 * 7 = 5040.Therefore, the number of possible four letter words is 5040. Verification / Alternative check:Using factorial form: 10P4 = 10! / 6!.10! = 3628800 and 6! = 720.Compute 3628800 / 720 = 5040, which matches the earlier result. Why Other Options Are Wrong:2525 and 2052 are not equal to 10P4 and result from incorrect multiplication or division.4521 is another random looking number that does not correspond to any standard combinatorial expression in this setup. Common Pitfalls:Using combinations 10C4 instead of permutations, which would ignore the importance of order.Accidentally allowing repetition of letters, which would change the counting logic completely.Forgetting that all letters of LOGARITHMS are distinct and trying to adjust for repeated letters unnecessarily. Final Answer:The total number of four letter words is 5040.

Question 14

In how many distinct ways can the letters of the word LEADER be arranged, given that the letter E appears twice?

Accepted Answer

Introduction / Context:This question tests your knowledge of permutations of multiset letters where some letters repeat. When a word contains repeated letters, we must adjust the standard factorial formula to avoid overcounting identical arrangements that arise from swapping identical letters. Given Data / Assumptions: The word is LEADER. Total letters = 6: L, E, A, D, E, R. The letter E appears twice, while L, A, D and R appear once each. We are asked to count all distinct permutations of these letters. Concept / Approach:If all letters were distinct, the number of arrangements would be 6! because there would be 6 positions and 6 distinct choices for each ordering. However, here E is repeated twice. Interchanging the two E letters does not create a new distinct word. To correct for overcounting, we divide by the factorial of the number of times each repeated letter appears. Thus the formula becomes total permutations = 6! / 2! for this word. Step-by-Step Solution:Total letters n = 6.If all were distinct, arrangements = 6! = 720.Letter E is repeated 2 times, so divide by 2! to account for identical swaps.Compute 2! = 2.Hence distinct arrangements = 6! / 2! = 720 / 2 = 360. Verification / Alternative check:There are no other repeated letters, so only one division factor (2!) is needed.You could manually check a smaller example such as a three letter word with two identical letters to see that division by 2! is correct in principle. Why Other Options Are Wrong:420 and 576 would require non integer division by a repetition factor or some incorrect counting logic.220 is too small and does not match any simple factorial division for this word. Common Pitfalls:Forgetting to divide by 2! for the repeated letter E, which would give 720 and double the correct answer.Dividing by an extra factorial such as 3! by incorrectly assuming another repeated letter.Confusing combinations with permutations and using 6C2 or similar expressions. Final Answer:The letters of LEADER can be arranged in 360 distinct ways.

Question 15

In how many different ways can the letters of the word CIRCUMSTANCES be arranged if each vowel (A, E, I, U) occupies an odd position and the letter N is fixed at the last (13th) position?

Accepted Answer

Introduction / Context:This question blends permutations with multiple constraints: fixed positions, repeated consonants and vowel placement restrictions. It is a good example of how to systematically handle non trivial word arrangement problems using combinatorics, especially when some letters must appear only at certain positions and there are repeated letters. Given Data / Assumptions: The word CIRCUMSTANCES has 13 letters. Letter counts: C appears 3 times, S appears 2 times, and I, R, U, M, T, A, N, E each appear once. Vowels are A, E, I and U (4 vowels). All vowels must be in odd positions. The letter N must always be at the last position, which is the 13th (an odd position). The remaining consonants fill the remaining positions. Concept / Approach:Positions 1 to 13 have 7 odd places: 1, 3, 5, 7, 9, 11 and 13. N is fixed at position 13, leaving 6 odd positions (1, 3, 5, 7, 9, 11) where the 4 distinct vowels must be placed. The 8 remaining consonant letters (C, C, C, S, S, R, M, T) then occupy the remaining 8 positions (2, 4, 6, 8, 10, 12 and the 2 unused odd positions). We count choices for vowel positions, permutations of vowels and permutations of consonants with repetition, then multiply all factors. Step-by-Step Solution:Odd positions available for vowels (excluding position 13): {1, 3, 5, 7, 9, 11}.Choose 4 of these 6 odd positions for the 4 vowels: number of ways = 6C4 = 15.Arrange the 4 distinct vowels (A, E, I, U) in the chosen positions: 4! = 24 ways.Remaining letters are consonants: C (3), S (2), R (1), M (1), T (1); total 8 consonants.Arrange these 8 letters in the remaining 8 positions. Because C repeats 3 times and S repeats 2 times, permutations = 8! / (3! * 2!).Compute 8! = 40320, 3! = 6 and 2! = 2, so denominator = 12, giving 40320 / 12 = 3360.Total arrangements = 15 * 24 * 3360.First, 15 * 24 = 360. Then 360 * 3360 = 1209600. Verification / Alternative check:You can re check multiplication: 3360 * 300 = 1008000 and 3360 * 60 = 201600, so sum = 1209600.There is no double counting because vowel positions and consonant positions are distinct and every step handles all symmetry from repeated letters. Why Other Options Are Wrong:151200 and 72000 are much smaller and typically arise if you ignore some positional choices or forget the division by 3! and 2! for repeated consonants.504020 does not align with any logically factored product related to this setup. Common Pitfalls:Placing vowels into all 7 odd positions instead of just 4, which does not reflect the actual number of vowels.Forgetting that N is fixed at the last position and accidentally treating it as a freely placeable consonant.Not accounting for the repeated consonants C and S, thereby overcounting arrangements. Final Answer:The required number of arrangements is 1209600.

Question 16

A box contains 5 green, 4 yellow and 3 white marbles. If three marbles are drawn at random, what is the probability that the three marbles are not all of the same colour?

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Introduction / Context:This question tests basic probability with combinations. Rather than directly counting the favourable cases, it is often easier to find the complement event (in this case, all marbles of the same colour) and subtract from 1. This technique simplifies calculations and is widely used in exam problems. Given Data / Assumptions: Green marbles: 5. Yellow marbles: 4. White marbles: 3. Total marbles: 5 + 4 + 3 = 12. We draw 3 marbles at random without replacement. We want the probability that the 3 drawn marbles are not all of the same colour. Concept / Approach:Probability = (number of favourable outcomes) / (total number of possible outcomes). The total number of ways to choose 3 marbles out of 12 is 12C3. The unfavourable case is when all 3 marbles are of the same colour (either all green, all yellow or all white). We first compute the probability of this unfavourable event and subtract it from 1 to get the required probability that the three are not all of the same colour. Step-by-Step Solution:Total ways to draw 3 marbles: 12C3 = 12 * 11 * 10 / 6 = 220.Ways to draw 3 green marbles: 5C3 = 10.Ways to draw 3 yellow marbles: 4C3 = 4.Ways to draw 3 white marbles: 3C3 = 1.Total ways with all same colour = 10 + 4 + 1 = 15.Probability(all same colour) = 15 / 220.Required probability = 1 - 15 / 220 = (220 - 15) / 220 = 205 / 220.Simplify the fraction by dividing numerator and denominator by 5: 205 / 220 = 41 / 44. Verification / Alternative check:You could also directly count all non same colour combinations, but that requires many cases.The complement method is shorter and less error prone, and the final simplified fraction 41 / 44 lies between 0 and 1 as expected. Why Other Options Are Wrong:40/44 is slightly smaller and would correspond to 200 favourable outcomes instead of 205.44/41 is greater than 1, which is impossible for a probability.40/39 is also greater than 1, so it cannot represent a probability. Common Pitfalls:Forgetting that there are three colours and only counting one all same colour case.Dividing by the wrong total number of outcomes or miscomputing 12C3.Not simplifying the final fraction when the question options are given in lowest terms. Final Answer:The probability that the three marbles are not all of the same colour is 41/44.

Question 17

An animal trainer wants to arrange 5 lions and 4 tigers in a single row so that no two lions stand next to each other. In how many distinct arrangements can this be done?

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Introduction / Context:This problem is a classic example of permutations with separation constraints, where certain objects (lions) are not allowed to be adjacent. You must use the idea of placing one type of animal first and then inserting the others into the available gaps to ensure the required separation condition is satisfied. Given Data / Assumptions: There are 5 lions and 4 tigers. All lions are considered distinct, and all tigers are considered distinct. The animals are to be arranged in a row. No two lions are allowed to stand next to each other. Concept / Approach:A standard method for such problems is to first arrange the animals that have no restriction between themselves, and then place the restricted animals (the lions) into the gaps created. Here, if we first arrange the 4 tigers, they will create 5 potential gaps for the lions: before the first tiger, between each pair of tigers, and after the last tiger. To avoid two lions being adjacent, we must place exactly one lion in each gap because we have exactly 5 lions and 5 gaps. Step-by-Step Solution:Step 1: Arrange the 4 distinct tigers in a row. Number of ways = 4! = 24.Step 2: The 4 tigers create 5 gaps: _ T _ T _ T _ T _.Step 3: Place one lion in each gap to ensure no two lions are adjacent. Since we have 5 lions and 5 gaps, each gap receives exactly one lion.Step 4: The 5 lions are distinct and can be arranged among these 5 positions in 5! = 120 ways.Step 5: Total valid arrangements = 4! * 5! = 24 * 120 = 2880. Verification / Alternative check:If you tried to start with placing lions, you would have to ensure tigers separate them, which is less straightforward.The gaps method automatically guarantees that no two lions stand together. Why Other Options Are Wrong:2800 and 2600 are close in magnitude but do not equal 4! * 5! and likely come from arithmetic errors.3980 is larger and does not correspond to any logical counting pattern in this setup. Common Pitfalls:Allowing two lions to occupy the same gap, which violates the non adjacency rule.Treating lions or tigers as indistinguishable when the problem context typically assumes them to be distinct individuals.Forgetting one of the outer gaps when counting possible positions for lions. Final Answer:The number of valid arrangements is 2880.

Question 18

From a group of 7 men and 6 women, in how many ways can a committee of 5 persons be formed such that at least 3 members of the committee are men?

Accepted Answer

Introduction / Context:This question is a standard committee selection problem involving combinations and a restriction on the minimum number of men. Such problems frequently appear in aptitude and competitive exams to test your ability to handle multiple cases and combine them correctly. Given Data / Assumptions: Total men available: 7. Total women available: 6. Committee size: 5 persons. Condition: At least 3 members must be men. Order of selection does not matter; only the combination of people matters. Concept / Approach:“At least 3 men” means the committee can have 3 men and 2 women, 4 men and 1 woman, or 5 men and 0 women. We calculate the number of committees for each case using combinations and then add the results. The general combination formula is nCr = n! / (r! * (n - r)!). Step-by-Step Solution:Case 1: 3 men and 2 women.Number of ways to choose men: 7C3 = 35.Number of ways to choose women: 6C2 = 15.Total for this case: 35 * 15 = 525.Case 2: 4 men and 1 woman.Ways to choose men: 7C4 = 35.Ways to choose women: 6C1 = 6.Total for this case: 35 * 6 = 210.Case 3: 5 men and 0 women.Ways to choose men: 7C5 = 21.Total for this case: 21.Now add all cases: 525 + 210 + 21 = 756. Verification / Alternative check:Total possible 5 person committees from 13 people is 13C5.You could compute committees with 0, 1 or 2 men and subtract from 13C5, but the direct case wise method above is simpler and less error prone here. Why Other Options Are Wrong:564, 735 and 657 arise from either omitting one of the required cases or miscomputing one of the combinations.None of these match the correct total of 756. Common Pitfalls:Misinterpreting “at least 3 men” as “exactly 3 men”.Forgetting to include the all men case (5 men and 0 women).Using permutations instead of combinations, which would incorrectly treat different orders as distinct committees. Final Answer:The number of ways to form such a committee is 756.

Question 19

A box contains 2 white balls, 3 black balls and 4 red balls. In how many different ways can 3 balls be drawn if at least one of the balls drawn must be black?

Accepted Answer

Introduction / Context:This question evaluates your understanding of combinations and the complement method in probability style counting. Instead of directly counting all selections that contain at least one black ball, it is easier to subtract the cases with no black balls from the total number of ways to draw three balls. Given Data / Assumptions: White balls = 2. Black balls = 3. Red balls = 4. Total balls = 2 + 3 + 4 = 9. We draw 3 balls at once. Condition: At least one ball among the three must be black. Concept / Approach:The total number of ways to choose 3 balls from 9 is 9C3. The only unwanted case is when none of the balls drawn is black. In that case, all three balls must come from the 2 white and 4 red balls, which is a total of 6 non black balls. So we calculate 6C3 and subtract it from 9C3 to get the desired count. Step-by-Step Solution:Step 1: Total ways to choose any 3 balls from 9 = 9C3.Compute 9C3 = 9 * 8 * 7 / 6 = 84.Step 2: Count ways with no black balls. Then we only use white and red balls: 2 white + 4 red = 6 balls.Number of ways to choose 3 non black balls = 6C3 = 6 * 5 * 4 / 6 = 20.Step 3: Favourable selections with at least one black ball = total selections - selections with no black ball.So required count = 84 - 20 = 64. Verification / Alternative check:You could also break into cases: exactly 1 black, exactly 2 black and exactly 3 black, and then add all counts.This alternative will also yield 64, but the complement method is more compact and avoids multiple case calculations. Why Other Options Are Wrong:48 and 45 are too small and can arise if one of the cases is missed.63 is just one less than the correct value and can result from an arithmetic slip such as mis computing 6C3. Common Pitfalls:Counting selections where at least one ball is black but forgetting to exclude the all non black scenario correctly.Miscomputing 9C3 or 6C3 due to factorial arithmetic errors.Mixing up order and combinations; here order does not matter, so combinations are correct. Final Answer:The number of ways to draw 3 balls with at least one black ball is 64.

Question 20

A standard deck of playing cards contains 13 spades. In how many distinct orders can these 13 spade cards alone be arranged in a row?

Accepted Answer

Introduction / Context:This question is a simple test of your understanding of permutations of distinct objects. Each spade card in a deck is different from the others, so arranging them in a sequence is a straightforward example of counting permutations with no repetition. Given Data / Assumptions: The deck has 13 different spade cards (for example, Ace of Spades, 2 of Spades, up to King of Spades). All 13 spade cards are distinct. We want to count the number of different orderings of these 13 cards in a row. Concept / Approach:When you arrange n distinct objects in a line and every arrangement is considered different, the number of possible arrangements is n factorial (n!). This arises because there are n choices for the first position, (n - 1) choices for the second, and so on, down to 1 choice for the last position. Multiplying these gives n!. Step-by-Step Solution:There are 13 distinct spade cards.The number of permutations of 13 distinct objects = 13! (13 factorial).13! means 13 * 12 * 11 * ... * 2 * 1.We do not need to compute the exact numerical value here because the expression 13! itself represents the correct count. Verification / Alternative check:If you wanted, you could compute the approximate size of 13! to get a sense of magnitude, but for exam purposes recognizing 13! as the permutation count is sufficient.No division by factorials is needed because there are no repeated cards. Why Other Options Are Wrong:13^2 (13 squared) counts something like choosing among 13 cards with replacement for only 2 positions, which is unrelated.13^13 represents a different model where each of 13 positions could take any of 13 cards with repetition allowed, which is not our situation.2! only counts the number of permutations of 2 items and is far too small. Common Pitfalls:Confusing exponent notation with factorial notation.Assuming some symmetry or repetition among cards that does not exist.Trying to divide by something because in many word problems letters repeat; here they do not. Final Answer:The number of distinct arrangements is 13!.

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