Question 1

Five distinct letters are to be posted into four distinct letter boxes. If each letter can go into any box independently of the others, in how many different ways can all five letters be posted?

Accepted Answer

Introduction / Context: This problem is about distributing distinct objects into distinct boxes when there is no restriction on how many objects each box can receive. Each of the five letters can be posted into any of the four letter boxes independently. Given Data / Assumptions: Number of letters = 5. Number of letter boxes = 4. Each letter can go into any of the 4 boxes. More than one letter can go into the same box. Letters and boxes are all distinct. Concept / Approach: For each letter, there are 4 possible choices of box. Since the placement of each letter is independent of the others, we can apply the multiplication rule of counting, raising the number of choices to a power equal to the number of letters. Step-by-Step Solution: Step 1: Focus on one letter at a time. For letter 1, there are 4 possible boxes. Step 2: For letter 2, again there are 4 possible boxes, regardless of where letter 1 went. Step 3: The same is true for each of the remaining letters; each has 4 choices of box. Step 4: Since these choices are independent, total number of ways = 4 * 4 * 4 * 4 * 4 = 4^5. Step 5: Compute 4^5 = 4 * 4 * 4 * 4 * 4 = 1024. Verification / Alternative check: You can view each distribution as a 5 digit code in base 4 where each position represents a letter and each digit represents the chosen box. The total number of such codes is 4^5, which again equals 1024. Why Other Options Are Wrong: 512 equals 2^9 or 8^3 and has no direct connection to the structure 4^5 in this context. 625 is 5^4, which corresponds to the reversed base. The value 20 would correspond to combinations or some limited case that does not fit this problem. Common Pitfalls: The main error is to treat letters as identical or to think in terms of combinations rather than independent placements. Some learners also mistakenly divide by factorials, which is not appropriate because different assignments of letters to boxes are distinct arrangements. Final Answer: There are 1024 different ways to post the five letters into the four letter boxes.

Question 2

Using the digits from 0 to 9, how many different 5 digit telephone numbers can be formed if any digit (including 0) is allowed in any position and repetition of digits is permitted?

Accepted Answer

Introduction / Context: This question is about counting the number of fixed length digit sequences formed from a given set of digits when repetition is allowed. We are asked how many 5 digit telephone numbers can be formed using digits from 0 to 9 with no extra restrictions. Given Data / Assumptions: Digits available: 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9. We need to form 5 digit numbers. Repetition of digits is allowed. A telephone number is considered valid even if it starts with 0. Concept / Approach: There are 10 possible digits for each position in the 5 digit number and each position choice is independent if repetition is allowed. Therefore, the total number of different 5 digit sequences is given by a power of 10. Step-by-Step Solution: Step 1: Consider the first digit. It can be any of the 10 digits from 0 to 9, so there are 10 possibilities. Step 2: The second digit can again be any of the 10 digits from 0 to 9. Step 3: The same reasoning applies to the third, fourth and fifth digits. Step 4: Using the multiplication rule of counting, total number of 5 digit numbers = 10 * 10 * 10 * 10 * 10 = 10^5. Step 5: Compute 10^5 = 100000, often written as 1,00,000. Verification / Alternative check: Each 5 digit number corresponds to a unique sequence of choices from the set of 10 digits. Since there are 10 possibilities for each of the 5 positions, the total of 10^5 combinations is logical and straightforward. Why Other Options Are Wrong: Values like 59,049 or 6561 are powers of 3 or combinations that arise in other contexts (for example, 3^10 or 3^8), and 3439 is a random value which does not match the structure of 10^5 for this problem. Common Pitfalls: Sometimes learners incorrectly impose a condition that the first digit cannot be 0, which would give 9 * 10^4 instead of 10^5. In this specific question, that extra restriction is not present because it explicitly refers to telephone numbers, which can be treated as fixed length sequences, not standard decimal integers. Final Answer: The number of possible 5 digit telephone numbers under these conditions is 1,00,000.

Question 3

You have 6 different New Year greeting cards and 4 friends. If each friend is to receive exactly one card, and you choose 4 of the 6 cards to send (one distinct card to each friend), in how many different ways can this be done?

Accepted Answer

Introduction / Context: This question is about distributing distinct objects to distinct recipients under the condition that each recipient receives exactly one object. We have 6 different greeting cards and 4 friends, and we send one card to each of the 4 friends, using 4 of the 6 cards. Given Data / Assumptions: Number of distinct greeting cards = 6. Number of friends = 4. Each friend must receive exactly one card. No friend receives more than one card. Cards are distinct and friends are distinct. Concept / Approach: We interpret the process in two steps. First, we select which 4 cards (out of 6) will be used. Second, we arrange those 4 selected cards among the 4 friends. Because order of assignment matters, we must count permutations rather than just combinations. Step-by-Step Solution: Step 1: Choose 4 cards from the 6 distinct cards. Number of ways = C(6, 4). Step 2: Evaluate C(6, 4) = 6 * 5 / (2 * 1) = 15. Step 3: For any fixed set of 4 cards, assign them to 4 friends. The number of ways to assign 4 distinct cards to 4 distinct friends is 4! = 24. Step 4: Total ways = number of ways to choose cards * number of ways to assign them = 15 * 24. Step 5: Compute 15 * 24 = 360. Verification / Alternative check: Another way is to view the process directly as permutations: we are creating an ordered selection of 4 cards from 6 for the 4 friends. This is 6P4 = 6 * 5 * 4 * 3, which equals 360 as well, confirming the two step reasoning. Why Other Options Are Wrong: 720 would correspond to 6!, which would be appropriate only if all 6 cards were used and arranged. 240 and 740 do not match the natural structure of 6P4 or C(6, 4) * 4! and therefore are miscalculations or guesses. Common Pitfalls: Students sometimes use only combinations and stop at C(6, 4) = 15, forgetting that each chosen set can be arranged among the friends in many ways. Others mistakenly use 6^4, which would allow a card to be sent to more than one friend, violating the condition that each card is used at most once. Final Answer: The number of ways to send 4 different cards to 4 friends, choosing from 6 available cards, is 360.

Question 4

A coach must choose a starting lineup of 5 players from a team of 12 players. In how many different ways can the coach select these 5 starters if only the choice of players matters and not their order?

Accepted Answer

Introduction / Context: This question involves combinations, where we are selecting a group of players and the order of selection does not matter. The coach needs to form a starting lineup of 5 players from a squad of 12. Given Data / Assumptions: Total players available = 12. Number of starters to be chosen = 5. Order of players in the list of starters does not matter; only the set of players is important. Concept / Approach: Since the order of selection does not matter, we use the combination formula nCr, which counts the number of ways to choose r objects from n distinct objects without regard to order. Here, n = 12 and r = 5. Step-by-Step Solution: Step 1: Use the combination formula: nCr = n! / (r! * (n - r)!). Step 2: Substitute n = 12 and r = 5. Step 3: Compute 12C5 = 12! / (5! * 7!). Step 4: Simplify by expanding the top 5 factors: 12C5 = (12 * 11 * 10 * 9 * 8) / (5 * 4 * 3 * 2 * 1). Step 5: Compute numerator: 12 * 11 = 132, 132 * 10 = 1320, 1320 * 9 = 11880, 11880 * 8 = 95040. Step 6: Compute denominator: 5! = 120. Step 7: Divide 95040 by 120 to obtain 792. Verification / Alternative check: You can check the arithmetic by simplifying before multiplying fully. For example, 10 / 5 = 2, 12 / 4 = 3, so we can compute (3 * 11 * 2 * 9 * 8) / (3 * 2 * 1) and obtain the same final result of 792, confirming the correctness of the combination calculation. Why Other Options Are Wrong: Values like 495 and 720 are common for related combinations (for example, 12C4 or 6! values) but do not match 12C5. 624 does not correspond to any standard selection count for this context. Common Pitfalls: A frequent mistake is to use permutations instead of combinations, which multiplies by unnecessary factors related to ordering. Another issue is miscalculating the factorial values or failing to simplify fractions properly before dividing. Final Answer: The number of different ways to choose 5 starters from 12 players is 792.

Question 5

If a, b and c are positive integers that satisfy a + b + c = 21, how many ordered triples (a, b, c) of positive integers are possible in total?

Accepted Answer

Introduction / Context: This is a classic problem of finding the number of positive integer solutions to a linear equation in several variables. We are given a + b + c = 21 with the condition that a, b and c are all positive integers, and we need to count how many ordered triples satisfy this equation. Given Data / Assumptions: a, b and c are positive integers (strictly greater than zero). a + b + c = 21. Different orders of the same numbers count as different solutions. Concept / Approach: This is a stars and bars problem. For positive integer solutions, the standard formula says that the number of solutions to a1 + a2 + ... + ak = n, with each ai greater than zero, is C(n - 1, k - 1). Here k = 3 and n = 21. Step-by-Step Solution: Step 1: Recognise that a, b and c are all strictly positive, which matches the positive integer stars and bars formula. Step 2: Use the formula for the number of positive integer solutions: C(n - 1, k - 1). Step 3: Substitute n = 21 and k = 3. Step 4: Compute C(21 - 1, 3 - 1) = C(20, 2). Step 5: Evaluate C(20, 2) = 20 * 19 / (2 * 1) = 190. Verification / Alternative check: You can transform variables by letting a = x + 1, b = y + 1 and c = z + 1 where x, y and z are non negative integers. Then x + y + z = 21 - 3 = 18. The number of non negative integer solutions of x + y + z = 18 is C(18 + 3 - 1, 3 - 1) = C(20, 2) = 190, confirming the result. Why Other Options Are Wrong: 109, 901 and 910 do not match the correct combinatorial formula for three positive integers summing to 21. They are not values of C(20, 2) or any simple binomial coefficient for this setup. Common Pitfalls: A common error is to use the formula for non negative solutions directly on 21, which would give C(21 + 3 - 1, 3 - 1) and an incorrect higher value. Another mistake is to forget that order matters, but in this formula ordered triples are already counted correctly. Final Answer: The total number of ordered positive integer triples (a, b, c) such that a + b + c = 21 is 190.

Question 6

Two cards are drawn one after another without replacement from a standard pack of 52 playing cards. What is the probability that both cards drawn are Aces?

Accepted Answer

Introduction / Context: This probability question involves drawing cards without replacement from a standard deck. We want the probability that both cards drawn are Aces, which requires sequential probabilities with a changing sample space after the first draw. Given Data / Assumptions: A standard deck has 52 cards. There are 4 Aces in the deck. Two cards are drawn one after another without replacement. We are interested in the event that both drawn cards are Aces. Concept / Approach: We use the multiplication rule for dependent events. The probability that the first card is an Ace is 4 out of 52. After drawing one Ace, there are 3 Aces left and only 51 cards remaining, so the probability that the second card is also an Ace is 3 out of 51. The probability that both events occur is the product of these two probabilities. Step-by-Step Solution: Step 1: Probability that the first card drawn is an Ace = 4 / 52. Step 2: After one Ace is drawn, remaining Aces = 3 and total remaining cards = 51. Step 3: Probability that the second card drawn is an Ace given the first was an Ace = 3 / 51. Step 4: Multiply to find the joint probability for both draws. P(both Aces) = (4 / 52) * (3 / 51). Step 5: Simplify: 4 / 52 = 1 / 13, and 3 / 51 = 1 / 17, since 51 = 3 * 17. Step 6: Therefore, P(both Aces) = (1 / 13) * (1 / 17) = 1 / 221. Verification / Alternative check: You can treat the sample space as ordered draws from 52 cards and confirm that the favourable outcomes are sequences of two Aces. There are 4 choices for the first Ace and 3 for the second, giving 12 sequences, while the total sequences of two cards is 52 * 51. The ratio 12 / (52 * 51) reduces to 1 / 221, consistent with the stepwise calculation. Why Other Options Are Wrong: 51/1221, 42/221 and 52/1245 do not match the exact product of the two conditional probabilities. They arise from incorrect simplification or from counting unordered outcomes instead of ordered draws. Common Pitfalls: Learners sometimes forget that the second draw is conditional on the first, and may treat probabilities as independent. Another mistake is to reduce fractions incorrectly or to mix up the number of remaining Aces and remaining cards after the first draw. Final Answer: The probability that both cards drawn are Aces is 1/221.

Question 7

From a group of 6 men and 9 women, in how many different ways can one couple be chosen, where each couple consists of one man and one woman?

Accepted Answer

Introduction / Context: This question involves forming a simple couple from two groups, one of men and one of women. A couple is defined here as one man together with one woman. We need to count how many different such couples can be formed. Given Data / Assumptions: Number of men = 6. Number of women = 9. Each couple consists of exactly one man and one woman. All men and women are distinct individuals. Concept / Approach: To form a couple, we independently choose one man from the 6 available and one woman from the 9 available. The total number of distinct couples is the product of the number of choices for the man and the number of choices for the woman. Step-by-Step Solution: Step 1: Choose a man from the 6 men. There are 6 possible choices. Step 2: For each fixed choice of man, choose a woman from the 9 women. There are 9 choices for the woman. Step 3: Use the multiplication rule of counting because the choice of man and choice of woman are independent decisions. Step 4: Total number of different couples = 6 * 9 = 54. Verification / Alternative check: You can think of listing pairs as (man, woman). For each of the 6 men, there is a row of 9 possible women, forming a grid of 6 by 9 = 54 unique ordered pairs. Each pair corresponds to a distinct couple, so the total is 54. Why Other Options Are Wrong: 26, 52 and 28 do not match the straightforward product 6 * 9. For example, 52 is close to 54 but indicates an incorrect subtraction or miscount of a small number of pairs. Common Pitfalls: A frequent error is to accidentally count only one side, such as just the men or just the women, or to use combinations in a way that mixes the two groups incorrectly. Another issue is confusing this simple counting with more complex pairings where multiple couples are chosen at once. Final Answer: The number of distinct man woman couples that can be formed from 6 men and 9 women is 54.

Question 8

The number of possible singing orders for 7 distinct players when the youngest player is not allowed to sing last

Accepted Answer

Introduction / Context: This question tests the concept of permutations with an additional restriction on the position of one particular player. Such questions are common in aptitude and placement exams because they combine basic factorial ideas with logical counting of restricted arrangements. Given Data / Assumptions: There are 7 distinct players who will sing one after another. One of these players is the youngest player. The youngest player must not be in the last position. All arrangements are simple linear orders with no ties and no repeats. Concept / Approach: Without restrictions, the number of ways to arrange n distinct people in a row is n factorial. When there is a restriction on the position of one person, we normally count total arrangements and subtract the arrangements that violate the restriction. This is called the complement method. Step-by-Step Solution: Step 1: Total unrestricted arrangements of 7 distinct players = 7!.Step 2: Compute 7! = 7 * 6 * 5 * 4 * 3 * 2 * 1 = 5040.Step 3: Count the arrangements where the youngest player is fixed in the last position.Step 4: If the youngest is last, the remaining 6 players can be arranged in the first 6 positions in 6! ways.Step 5: Compute 6! = 6 * 5 * 4 * 3 * 2 * 1 = 720.Step 6: The number of valid arrangements is total arrangements minus invalid arrangements.Step 7: Valid arrangements = 7! - 6! = 5040 - 720 = 4320. Verification / Alternative check: You can also think of this as placing the youngest in any of the first 6 positions. First choose a position for the youngest in 6 ways, then arrange the remaining 6 players in the remaining 6 positions in 6! ways. So total valid arrangements = 6 * 6! = 6 * 720 = 4320, which matches the previous calculation. Why Other Options Are Wrong: 2580 is much smaller than 4320 and does not correspond to any standard factorial based count for 7 people. 3687 is not close to any natural expression like 7!, 7! - 6! or 6 * 6! and seems arbitrary. 5460 is larger than the total number of possible permutations 7!, so it is impossible. Common Pitfalls: Candidates often simply calculate 7! and stop, forgetting the restriction on the youngest player. Another common error is to try to directly count arrangements where the youngest is in allowed positions but then forget to multiply by all permutations of the remaining players. The safest method is to use the complement approach: count all arrangements and subtract the ones that violate the condition. Final Answer: The youngest player can sing in 4320 different valid orders, so the correct answer is 4320.

Question 9

Number of distinct arrangements of the word SCOOTY when letters S and Y are fixed at the two ends

Accepted Answer

Introduction / Context: This problem checks your understanding of permutations of letters in a word when certain positions are fixed. The word SCOOTY contains repeating letters, so you must apply the permutation formula for objects with repetition, combined with positional restrictions. Given Data / Assumptions: Word: SCOOTY. Letters: S, C, O, O, T, Y. The letters S and Y must always occupy the two end positions in some order. The remaining letters C, O, O, T fill the middle four positions. All arrangements are case insensitive and every arrangement counts as a distinct word even if it has no meaning. Concept / Approach: We fix which letters go at the ends and then arrange the remaining letters in the middle positions. Because there is a repeated letter O, we use the formula for permutations of n objects with repetition: n factorial divided by the product of factorials of the counts of identical objects. Step-by-Step Solution: Step 1: The two ends must be S and Y. They can appear as S at the left and Y at the right or Y at the left and S at the right.Step 2: Number of ways to arrange S and Y at the two ends = 2! = 2.Step 3: The middle four positions must be filled with C, O, O, T.Step 4: If all four letters were distinct, we would have 4! arrangements.Step 5: But O appears twice, so the count must be divided by 2! to correct for repetition.Step 6: Number of arrangements of C, O, O, T = 4! / 2! = 24 / 2 = 12.Step 7: Total valid arrangements = ways to arrange the ends * ways to arrange the middle = 2 * 12 = 24. Verification / Alternative check: You can explicitly imagine fixing S on the left and Y on the right, then list or count the permutations of C, O, O, T to confirm there are 12. Repeating the same with Y on the left and S on the right would again give 12, so the total is 24. This confirms the formula based approach. Why Other Options Are Wrong: 720 would be the number of arrangements of 6 distinct letters, which does not apply due to repetition and fixed ends. 360 is still too large and does not match 2 * 4! or any correct adjustment for repeated letters. 120 equals 5! and would correspond to an incorrect assumption about only 5 letters being arranged freely. Common Pitfalls: Students often ignore the repeated letter O and simply compute 2 * 4!, giving 48, which is double the correct answer. Sometimes, they may also forget that S and Y can swap ends, leading them to count only 12 arrangements. Always separate the problem into fixed positions and free positions and then apply the permutation rules carefully, especially when letters repeat. Final Answer: The number of valid arrangements is 24, so the correct answer is 24.

Question 10

Number of ways to draw 3 balls from a box with at least one green ball when the box has 2 blue, 3 green and 4 yellow balls

Accepted Answer

Introduction / Context: This question involves combinations with a restriction, which is a very common type in aptitude exams. You must count the number of ways to choose 3 balls from a set of coloured balls, under the condition that at least one of the chosen balls is green. Given Data / Assumptions: The box contains 2 blue balls. The box contains 3 green balls. The box contains 4 yellow balls. All balls of the same colour are identical in colour but considered distinct objects for counting combinations. We must select exactly 3 balls. The selection must include at least one green ball. Concept / Approach: A standard strategy for at least one type problems is to first count all possible selections without any restriction and then subtract the selections that violate the condition. Here, the undesired selections are those that contain no green balls. We use the combination formula nCr to count selections. Step-by-Step Solution: Step 1: Total number of balls = 2 blue + 3 green + 4 yellow = 9.Step 2: Total ways to choose any 3 balls from 9 balls = 9C3.Step 3: Compute 9C3 = 9 * 8 * 7 / (3 * 2 * 1) = 84.Step 4: Now count the selections with no green ball at all.Step 5: Balls that are not green are 2 blue + 4 yellow = 6 balls.Step 6: Number of ways to choose 3 balls all from these 6 non green balls = 6C3 = 6 * 5 * 4 / 6 = 20.Step 7: Valid selections with at least one green ball = total selections minus selections with no green = 84 - 20 = 64. Verification / Alternative check: You could also directly split into cases: exactly 1 green, exactly 2 green, and exactly 3 green, and compute each using combinations. Adding those case counts should again give 64. If you do this and the sum matches 64, it confirms that the complement method is correct. Why Other Options Are Wrong: 48 and 32 are smaller numbers that might arise from partial case counting but not from a complete and correct calculation. 24 is far too small and ignores many valid combinations. Common Pitfalls: Many students mistakenly treat colours as identical groups and do not count distinct balls separately, which leads to undercounting. Another typical error is to count only one or two cases, such as exactly one green ball, without including other valid cases. The complement method is usually easier and less error prone for at least one type questions. Final Answer: The number of ways to draw 3 balls including at least one green ball is 64, so the correct answer is 64.

Question 11

Number of ways to select one or more balls from 6 identical green balls and 4 identical blue balls

Accepted Answer

Introduction / Context: This question checks your understanding of combinations with repeated identical objects. The balls of the same colour are identical, so the selection is determined only by how many of each colour you choose, not which specific ball you pick. The task is to count all selections that are not empty. Given Data / Assumptions: There are 6 identical green balls. There are 4 identical blue balls. A selection is described only by the numbers of green and blue balls chosen. We must choose at least one ball in total. Order of selection does not matter; only the counts matter. Concept / Approach: We model the selection by a pair of integers (g, b) where g is the number of green balls chosen and b is the number of blue balls chosen. Valid values are 0 to the available count for each colour. We count all possible pairs except the pair where both counts are zero, because that pair corresponds to choosing no balls. Step-by-Step Solution: Step 1: Possible numbers of green balls g can range from 0 to 6, inclusive, so there are 7 possible values for g.Step 2: Possible numbers of blue balls b can range from 0 to 4, inclusive, so there are 5 possible values for b.Step 3: If there were no restriction, the total number of (g, b) pairs would be 7 * 5 = 35.Step 4: Among these 35 pairs, the pair (0, 0) represents choosing no green and no blue balls, that is, an empty selection.Step 5: The question asks for one or more balls, so the empty selection must be excluded.Step 6: Valid selections = total pairs minus empty pair = 35 - 1 = 34. Verification / Alternative check: You can also explicitly count non empty selections by considering g from 0 to 6 and for each g count allowed b values such that g and b are not both zero. Summing these counts will also give 34. This confirms the product minus one method. Why Other Options Are Wrong: 42 would correspond to some incorrect counting, possibly treating balls as distinct rather than identical. 31 and 22 are underestimates that suggest missed combinations, perhaps by forcing at least one of each colour, which is not required. Common Pitfalls: A common mistake is to think in terms of distinct balls and try to use standard combinations, which is not appropriate when balls are identical in each colour. Another frequent error is forgetting to subtract the case where no balls are selected. Always remember that one or more means every non empty selection is allowed. Final Answer: The total number of ways to choose one or more balls is 34, so the correct answer is 34.

Question 12

Number of ways 10 speakers A1 to A10 can speak when A1 must always speak after A2

Accepted Answer

Introduction / Context: This question is a classic example of permutations with an order restriction between two particular elements. You are given 10 speakers and a condition that one specific speaker must speak after another specific speaker. Such problems help build strong intuition for symmetry and counting methods. Given Data / Assumptions: Speakers are labeled A1, A2, A3, up to A10. All speakers are distinct. All 10 speakers must speak exactly once. A1 must speak after A2 in the final order. No other restrictions are imposed. Concept / Approach: When there is a simple condition like A1 must come after A2 and there are no other constraints, we can use a symmetry argument. In all possible permutations of 10 distinct speakers, A1 and A2 are equally likely to appear in any relative order. Therefore exactly half of all permutations will satisfy A1 after A2 and the other half will have A1 before A2. Step-by-Step Solution: Step 1: Count total unrestricted permutations of 10 distinct speakers.Step 2: Total permutations = 10!.Step 3: Consider the pair (A1, A2). In any particular permutation, either A1 comes before A2 or A2 comes before A1.Step 4: There is perfect symmetry between these two possibilities, so the number of permutations where A1 is before A2 equals the number where A1 is after A2.Step 5: Therefore, the number of permutations with A1 after A2 is exactly half of all permutations.Step 6: Required count = 10! / 2. Verification / Alternative check: Another way is to think of first arranging all 10 speakers in 10! ways, then looking at each arrangement in pairs. If you swap the positions of A1 and A2 in any arrangement where A1 is before A2, you get a unique arrangement where A1 is after A2, and vice versa. This one to one correspondence confirms that the count is exactly half of 10!. Why Other Options Are Wrong: 9! and 9!/2 ignore that all 10 speakers are being permuted, not 9. 10! counts all permutations without applying the restriction that A1 must be after A2. Common Pitfalls: Students sometimes attempt to count positions directly, such as choosing a position for A2 and later choosing a position for A1, which often leads to double counting or missing cases. The symmetry argument is both simpler and more reliable for this type of relative order condition. Final Answer: The number of valid speaking orders is 10! divided by 2, so the correct answer is 10! / 2.

Question 13

Number of arrangements of the word CAPITAL so that all vowels remain together in every arrangement

Accepted Answer

Introduction / Context: This problem focuses on permutations of letters in a word with the additional condition that all vowels must appear together as one block. It combines the idea of treating a group of letters as a single unit and then arranging letters inside that unit. Given Data / Assumptions: Word: CAPITAL. Letters: C, A, P, I, T, A, L. Vowels: A, I, A. Consonants: C, P, T, L. All vowels must appear together in every arrangement. Letters A appear twice and are identical. Concept / Approach: First we treat the group of vowels as a single block so that they always remain together. Then we arrange this block together with the consonants. Inside the vowel block, we also need to count the permutations of the vowels, taking into account the repeated letter A. Step-by-Step Solution: Step 1: Group the three vowels A, A, I as a single block, say V.Step 2: Now we have the block V and the four consonants C, P, T, L.Step 3: So we have 5 items in total: V, C, P, T, L.Step 4: These 5 distinct items can be arranged in 5! ways, so we have 5! = 120 arrangements of the block and consonants.Step 5: Inside the vowel block V, the letters are A, A, I.Step 6: If they were all distinct, we would have 3! permutations, but the letter A is repeated twice.Step 7: Number of permutations inside the block = 3! / 2! = 6 / 2 = 3.Step 8: Total valid arrangements = arrangements of blocks * internal vowel arrangements = 120 * 3 = 360. Verification / Alternative check: You can cross check by listing the three internal vowel arrangements, such as A A I, A I A, and I A A, and then imagine placing this block in 5! ways among the consonants. Since each internal arrangement can be paired with 120 external arrangements, the total must be 360, which confirms our calculation. Why Other Options Are Wrong: 720 would correspond to 6! for some miscounted set, which ignores repetition of A and the block idea. 120 counts only the arrangements of the block and consonants without considering internal vowel permutations. 840 does not match any standard factorial based combination of these counts. Common Pitfalls: Students often forget to divide by 2! for the repeated A, leading to 5! * 3! which equals 720. Another error is to arrange all 7 letters first and then try to subtract cases where vowels are not together, which becomes complicated. Treating the vowels as one block from the start is the clean and simple method. Final Answer: The number of valid arrangements of CAPITAL with all vowels together is 360, so the correct answer is 360.

Question 14

Number of arrangements of the letters of the word BENGALI when all three vowels always appear together

Accepted Answer

Introduction / Context: This question involves permutations of the letters of a word with the condition that all vowels must be grouped together. Such problems test your ability to use the block method and count permutations when there are both consonants and vowels. Given Data / Assumptions: Word: BENGALI. Letters: B, E, N, G, A, L, I. All seven letters are distinct. Vowels: E, A, I. Consonants: B, N, G, L. The three vowels must always appear together as one group in every arrangement. Concept / Approach: We treat the group of vowels as a single block so that they stay together. Then we arrange this block along with the consonants. Finally, we count the internal arrangements of the vowels inside their block and multiply the two counts. Step-by-Step Solution: Step 1: Group the vowels E, A, I into a single block V.Step 2: The consonants are B, N, G, L, which are four distinct letters.Step 3: Now we have five items to arrange: V, B, N, G, L.Step 4: Number of permutations of these five distinct items = 5! = 120.Step 5: Inside the vowel block V, the letters are E, A, I, all distinct.Step 6: These three vowels can be arranged among themselves in 3! = 6 ways.Step 7: Total arrangements with vowels always together = 5! * 3! = 120 * 6 = 720. Verification / Alternative check: If you wish, you can think of first arranging the four consonants in 4! ways and then inserting the vowel block V in any of the five possible slots around these consonants. That again gives 4! * 5 * 3! = 24 * 5 * 6 = 720, which matches the earlier calculation, because 5! = 4! * 5. Why Other Options Are Wrong: 120 counts only the arrangements of the five items V, B, N, G, L and ignores internal vowel permutations. 640 and 540 do not align with any standard factorial or block count, so they are incorrect. Common Pitfalls: Students sometimes misread the condition and try to count arrangements where no two vowels are together instead of all vowels together. Others forget to account for the internal arrangements of vowels and stop at 5!. When dealing with block problems, always consider both the outer arrangement of blocks and the inner arrangement within each block. Final Answer: The number of arrangements of BENGALI with all vowels together is 720, so the correct answer is 720.

Question 15

Number of ways to arrange 4 novels and 2 biographies on a shelf chosen from 5 novels and 4 biographies

Accepted Answer

Introduction / Context: This problem mixes combinations and permutations. You must first choose which books are part of the arrangement and then consider in how many ways they can be placed in order on the shelf. This is a common pattern in counting problems. Given Data / Assumptions: There are 5 distinct novels. There are 4 distinct biographies. We want to select 4 novels and 2 biographies. The selected books are then arranged in a row on a shelf. Order of books on the shelf matters. Concept / Approach: The process has two stages. First we select the required books using combinations, because the selection itself does not care about order. After selection, we arrange the chosen books in all possible orders using permutations, because order on the shelf matters. Step-by-Step Solution: Step 1: Choose 4 novels out of 5 distinct novels. This can be done in 5C4 ways.Step 2: Compute 5C4 = 5.Step 3: Choose 2 biographies out of 4 distinct biographies. This can be done in 4C2 ways.Step 4: Compute 4C2 = 6.Step 5: Total ways to select the required 6 books (4 novels and 2 biographies) = 5C4 * 4C2 = 5 * 6 = 30.Step 6: After selection, we have 6 distinct books to place on the shelf.Step 7: Number of permutations of 6 distinct books = 6! = 720.Step 8: Total arrangements = number of selections * arrangements of each selection = 30 * 720 = 21600. Verification / Alternative check: You can sanity check by noting that if all 9 books were distinct and all 6 positions were free, the total permutations would be 9P6, which is larger than any option. Our value 21600 is noticeably smaller than 9P6 and consistent with first selecting, then arranging. Why Other Options Are Wrong: 26100, 24000, and 36000 are nearby large numbers but do not match the exact computation of 30 * 720. They may result from mistakes such as using 9C6 instead of the correct selection structure or miscalculating factorial values. Common Pitfalls: Candidates often confuse combinations and permutations. Some simply compute 9C6 and then multiply by 6!, forgetting the selection constraint of exactly 4 novels and exactly 2 biographies. Others might add counts instead of multiplying them. Always separate the problem into selection and arrangement phases and handle each carefully. Final Answer: The number of ways to select and arrange 4 novels and 2 biographies is 21600, so the correct answer is 21600.

Question 16

Number of different arrangements of the word HALLUCINATION when all consonants must appear together as one group

Accepted Answer

Introduction / Context: This problem involves counting arrangements of a long word with repeated letters and a special condition that all consonants must appear together as a single block. It combines several concepts: identification of vowels and consonants, handling repeated letters, and the block method. Given Data / Assumptions: Word: HALLUCINATION. Letters and counts: A (2), L (2), I (2), N (2), H (1), U (1), C (1), T (1), O (1). Total letters: 12. Vowels: A, A, U, I, I, O (6 letters). Consonants: H, L, L, C, N, N, T (7 letters). All consonants must appear together as one block. Concept / Approach: First, treat all 7 consonants as a single block. Then arrange this block together with the 6 vowel letters, taking care of repeated letters. Next, count the internal arrangements of the consonant block itself, again handling repetitions. Finally, multiply the two counts. Step-by-Step Solution: Step 1: Form a consonant block C that contains H, L, L, C, N, N, T.Step 2: Outside this block, we have 6 vowels A, A, U, I, I, O.Step 3: So we now have 7 items to arrange: C and the 6 vowel letters.Step 4: Among these 7 items, vowels A and I are repeated twice each, while C is distinct as a block.Step 5: Number of ways to arrange C, A, A, U, I, I, O = 7! / (2! * 2!) = 5040 / 4 = 1260.Step 6: Now consider the internal arrangements of the consonant block C which has H, L, L, C, N, N, T.Step 7: This is 7 letters with L repeated twice and N repeated twice, so arrangements = 7! / (2! * 2!).Step 8: Compute 7! / (2! * 2!) = 5040 / 4 = 1260.Step 9: Total arrangements with all consonants together = 1260 * 1260 = 1587600. Verification / Alternative check: You can sanity check the magnitude: without any restriction, total arrangements of 12 letters with these repetitions would be 12! divided by the product of factorials of the repeated counts. The answer 1587600 is less than that total, as it should be, because we are imposing a restriction. The answer also matches 1260 squared, confirming the two stage block reasoning. Why Other Options Are Wrong: 129780 and 35600 are much smaller than the correct count and arise from partial counting that misses either the internal or external arrangements. None of these is incorrect because option 1587600 already matches the correct computation. Common Pitfalls: Students often ignore repetition and simply use 7! or 12! leading to large incorrect numbers. Another mistake is to count only the arrangement of the block with vowels and forget to multiply by the internal arrangements of consonants. Always identify repeated letters clearly and divide by their factorials in every permutation stage. Final Answer: The number of valid arrangements is 1587600, so the correct answer is 1587600.

Question 17

Number of non negative integer solutions of a + b + c = 21

Accepted Answer

Introduction / Context: This question is a standard example of integer solution counting using the stars and bars method. You are asked to count the number of non negative integer solutions of a linear equation with three variables, which is a staple problem in combinatorics. Given Data / Assumptions: Equation: a + b + c = 21. Variables a, b, and c are non negative integers. Non negative means each variable can be zero or any positive integer as long as the sum is 21. Order of the triple (a, b, c) matters, so different assignments count as different solutions. Concept / Approach: The stars and bars method is used to count solutions of x1 + x2 + ... + xk = n where all xi are non negative integers. The formula for the number of solutions is n plus k minus 1 choose k minus 1. Here n is the total sum and k is the number of variables. Step-by-Step Solution: Step 1: Identify the total sum n = 21.Step 2: The number of variables is k = 3 corresponding to a, b, and c.Step 3: For non negative integer solutions, the formula is (n + k - 1)C(k - 1).Step 4: Substitute n = 21 and k = 3 into the formula.Step 5: We get (21 + 3 - 1)C(3 - 1) = 23C2.Step 6: Compute 23C2 = 23 * 22 / 2 = 253.Step 7: Thus there are 253 distinct non negative integer solutions for the equation. Verification / Alternative check: You can verify the formula by checking a simpler case such as a + b + c = 2. There you could list all solutions and confirm that the stars and bars formula predicts the correct count. Once you trust the formula, applying it to 21 is straightforward. Why Other Options Are Wrong: 123, 321, and 231 are close in size to 253 but none equals the correct binomial coefficient 23C2. They likely arise from misusing the formula such as 21C2 or 22C2. Common Pitfalls: A major source of error is confusing non negative and positive solutions. For strictly positive solutions, the formula is (n - 1)C(k - 1), which would give a different answer. Another mistake is plugging n and k incorrectly, for example computing (n + k)Ck. Always write the formula clearly and substitute values carefully. Final Answer: The total number of non negative integer solutions of a + b + c = 21 is 253, so the correct answer is 253.

Question 18

Number of distinct flag signals using 5 different coloured flags when any non empty selection and order is allowed

Accepted Answer

Introduction / Context: This question models signal making with different coloured flags. Each signal is formed by hoisting some flags in a particular order. The key idea is that any number of flags from one to all five may be used, and no flag is repeated within one signal. Given Data / Assumptions: There are 5 flags, each of a different colour. Any number of flags can be hoisted at a time, from 1 to 5. Within a signal, flags are arranged in order, so sequence matters. No flag is repeated in a single signal. Signals with different orders or different chosen flags are considered distinct. Concept / Approach: For a fixed number r of flags, the number of ordered arrangements of r flags chosen from 5 distinct flags is given by permutations, written as 5Pr. Since r can be 1, 2, 3, 4, or 5, we sum these permutation counts over all possible r. Step-by-Step Solution: Step 1: For r = 1, the number of signals is 5P1 = 5.Step 2: For r = 2, the number of signals is 5P2 = 5 * 4 = 20.Step 3: For r = 3, the number of signals is 5P3 = 5 * 4 * 3 = 60.Step 4: For r = 4, the number of signals is 5P4 = 5 * 4 * 3 * 2 = 120.Step 5: For r = 5, the number of signals is 5P5 = 5! = 120.Step 6: Sum all signals: 5 + 20 + 60 + 120 + 120 = 325. Verification / Alternative check: Notice that 5P1 through 5P5 are standard permutation values and the sum 325 is reasonable given the rapid growth of permutations. There is no simpler closed form for this sum here, so direct addition is the efficient and accurate method. Why Other Options Are Wrong: 235 and 253 are smaller than 325 and may result from omitting some terms, for example summing only up to 5P4. None of these is incorrect because option 325 exactly matches the correct total. Common Pitfalls: Students sometimes confuse this with combination counting and use 2^5 minus 1, which counts unordered selections, not signals where order matters. Another mistake is assuming only one flag can be hoisted, leading to a count of 5. Always check whether the order of flags in a signal changes its meaning. Final Answer: The number of distinct flag signals that can be generated is 325, so the correct answer is 325.

Question 19

Number of handshakes when every person in a group of 23 people shakes hands with every other person exactly once

Accepted Answer

Introduction / Context: This is a classic handshake problem that tests your understanding of combinations. Each handshake involves a pair of distinct people, and every pair shakes hands exactly once. You must avoid double counting pairs. Given Data / Assumptions: There are 23 people in a group. Each pair of distinct people shakes hands exactly once. No person shakes hands with themselves. The order inside a pair does not matter, so person A shaking hands with person B is the same as person B shaking hands with person A. Concept / Approach: Each handshake corresponds to a unique unordered pair of people. When there are n people, the number of unordered pairs is given by the combination nC2. This is because we must choose 2 distinct people out of n without considering order. Step-by-Step Solution: Step 1: Let n = 23, the total number of people.Step 2: The number of distinct handshakes equals the number of ways to choose 2 people from 23.Step 3: Use the formula for combinations: nC2 = n * (n - 1) / 2.Step 4: Substitute n = 23 to get 23C2 = 23 * 22 / 2.Step 5: Compute 23 * 22 = 506 and then divide by 2 to get 506 / 2 = 253.Step 6: Therefore, total handshakes = 253. Verification / Alternative check: Another way to see this is to imagine that each of the 23 people shakes hands with the remaining 22 people. This gives 23 * 22 handshake actions, but each handshake is counted twice (once from each person), so you divide by 2. That again gives 23 * 22 / 2 = 253. Why Other Options Are Wrong: 142, 175, and 212 are smaller than 253 and do not correspond to the correct combination computation. They may result from arithmetic mistakes or attempts to subtract extra counts incorrectly. Common Pitfalls: Some learners mistakenly use permutations and compute 23P2, which counts ordered pairs and gives a much larger number. Others might forget to divide by 2 when using the product method. Always remember that each handshake is an unordered pair and that nC2 is the correct formula. Final Answer: The total number of handshakes in the group is 253, so the correct answer is 253.

Question 20

Number of different ways to choose questions when a paper has 3 sections of 5 questions each and a student must attempt 4 questions from each section

Accepted Answer

Introduction / Context: This question is about choosing questions from a structured paper. There are several sections, each with a fixed number of questions, and the student must attempt a given number from each section. The task is to count the number of different sets of questions the student can select. Given Data / Assumptions: The question paper has 3 sections. Each section contains 5 distinct questions. The student must select exactly 4 questions from each section. The order in which the student answers questions does not matter; only which questions are chosen matters. Concept / Approach: The choice in each section is independent of the others. In each section, the student chooses 4 out of 5 questions, so the count for one section is a combination. To find the total number of ways for all sections together, we multiply the number of choices for each section. Step-by-Step Solution: Step 1: Consider a single section with 5 questions.Step 2: The number of ways to choose 4 questions from 5 is 5C4.Step 3: Compute 5C4 = 5.Step 4: Since there are 3 sections, each with 5 questions, and the student chooses 4 from each, the same count 5 applies to every section.Step 5: Choices across sections are independent, so total ways = 5 * 5 * 5.Step 6: Compute 5 * 5 * 5 = 125. Verification / Alternative check: You can think of the total selection as specifying which question is omitted in each section. In each section the student omits exactly 1 question, which can be chosen in 5 ways. Therefore, for 3 sections, the number of ways to select which question is omitted from each section is 5^3 = 125, matching the earlier calculation. Why Other Options Are Wrong: 1000, 625, and 525 are not equal to 5^3 and do not arise from correctly applying the combination and multiplication rules. For instance, 625 equals 5^4 and would correspond to an incorrect assumption about four independent choices of 5 options. Common Pitfalls: One common mistake is to add the choices for each section instead of multiplying them. Another is to confuse the required number of questions per section and attempt to choose 4 questions total across all sections. Always carefully interpret whether the conditions apply separately to each section or to the whole paper. Final Answer: The student can select questions in 125 different ways, so the correct answer is 125.

Permutation and Combination Questions