Q1. In how many different ways can the letters of the word 'COMMITTEE' be arranged?Correct Option C … Explanation: 'COMMITTEE' has 9 letters in which M, T and E each occur twice. Number of arrangements = 9! / (2! × 2! × 2!) = 362880 / 8 = 45360. Option A (90720) divides by only two of the repeated letters, and Option D (181440) divides by only one, so both overcount. Option E (22680) divides by an extra factor of 2. Option B (60480) does not correspond to the correct treatment of the repetitions. Hence, option C.Q2. In how many different ways can the letters of the word 'DAUGHTER' be arranged so that the word starts and ends with a vowel?Correct Option D … Explanation: 'DAUGHTER' has 8 distinct letters, of which 3 are vowels (A, U, E). The first and last places can be filled with vowels in 3 × 2 = 6 ways. The remaining 6 letters can be arranged in the middle 6 places in 6! = 720 ways. Total = 6 × 720 = 4320. Option A (5040) is simply 7!, which ignores the vowel restriction at both ends. The other options do not match the correct product. Hence, option D.Q3. How many even four-digit numbers can be formed from the digits 0, 1, 2, 3, 4, 5, 6 without repetition of digits?Correct Option E … Explanation: Case 1: units digit is 0. The remaining three places can be filled in 6 × 5 × 4 = 120 ways. Case 2: units digit is 2, 4 or 6 (3 choices). The thousands digit cannot be 0 or the units digit, leaving 5 choices; the hundreds and tens places can then be filled in 5 × 4 = 20 ways. This gives 3 × 5 × 20 = 300 numbers. Total = 120 + 300 = 420. Option B (300) counts only Case 2 and misses the numbers ending in 0. Hence, option E.Q4. How many five-digit numbers can be formed from the digits 1, 2, 3, 4, 5, 6, 7 without repetition such that the digit in the ten-thousands place is smaller than the digit in the thousands place?Correct Option C … Explanation: Total five-digit numbers without repetition = 7P5 = 7 × 6 × 5 × 4 × 3 = 2520. Since the two leading digits are always different, in exactly half of these arrangements the ten-thousands digit is smaller than the thousands digit, and in the other half it is larger. Required count = 2520 / 2 = 1260. Option A (2520) is the total count without applying the condition. Hence, option C.Q5. In how many ways can 8 students be arranged in a row so that two particular students, Krishna and Meera, are not next to each other?Correct Option E … Explanation: Total arrangements of 8 students = 8! = 40320. Arrangements where Krishna and Meera sit together: treat them as one unit, giving 7! × 2! = 5040 × 2 = 10080. Required arrangements = 40320 − 10080 = 30240. Option A (10080) is the number of arrangements in which they ARE together. Option D (35280) subtracts only 7! and forgets that the two students can swap places within their unit. Hence, option E.Q6. In how many ways can 5 men and 3 women be arranged in a row so that no two women are adjacent?Correct Option C … Explanation: First arrange the 5 men in 5! = 120 ways. This creates 6 gaps (including both ends) where women can stand without being adjacent. The 3 women can be placed in these 6 gaps in 6P3 = 6 × 5 × 4 = 120 ways. Total = 120 × 120 = 14400. Option A (7200) uses 6C3 × 3! × 5! / 2, halving the result incorrectly, while the other options arise from miscounting the gaps. Hence, option C.Q7. How many five-letter words, with or without meaning, containing exactly 3 vowels and 2 consonants can be formed from the letters of the word 'EDUCATION'?Correct Option B … Explanation: 'EDUCATION' has 9 distinct letters: 5 vowels (E, U, A, I, O) and 4 consonants (D, C, T, N). Choose 3 vowels in 5C3 = 10 ways and 2 consonants in 4C2 = 6 ways. Each selection of 5 letters can be arranged in 5! = 120 ways. Total = 10 × 6 × 120 = 7200. Option A (3600) is half the correct value, and the remaining options do not follow from the correct selection-and-arrangement product. Hence, option B.Q8. A committee of 5 is to be formed from 6 men and 4 women such that the committee has more men than women. In how many ways can this be done?Correct Option D … Explanation: More men than women in a committee of 5 means 3M + 2W, 4M + 1W or 5M + 0W. 3M + 2W: 6C3 × 4C2 = 20 × 6 = 120. 4M + 1W: 6C4 × 4C1 = 15 × 4 = 60. 5M: 6C5 = 6. Total = 120 + 60 + 6 = 186. Option C (180) forgets the all-men committee. Hence, option D.Q9. A bag contains 3 red, 4 blue and 5 green balls. In how many ways can 4 balls be drawn if at least one red ball must be included?Correct Option B … Explanation: Total balls = 12. Total ways to draw 4 balls = 12C4 = 495. Ways with no red ball (drawing from the 9 non-red balls) = 9C4 = 126. Ways with at least one red = 495 − 126 = 369. Option C (495) is the total without any condition, and Option E (126) is the number of draws with no red ball. Hence, option B.Q10. From 14 players, 7 are to be selected for a tournament. In how many ways can the selection be made if 3 particular players are always included and 2 particular players are always excluded?Correct Option A … Explanation: With 3 players always included, only 7 − 3 = 4 more players need to be chosen. With 3 included and 2 excluded, the pool left is 14 − 3 − 2 = 9 players. Required ways = 9C4 = 126. Option B (84) is 9C3, which picks the wrong number of remaining players, and Option C (210) is 10C4, which forgets one of the excluded players. Hence, option A.Q11. A quiz has 4 questions, and each question has 5 options. A candidate may attempt any question by choosing exactly one option, or leave it unattempted. In how many ways can the candidate attempt one or more questions?Correct Option D … Explanation: Each question can be handled in 6 ways: any one of the 5 options, or left unattempted. For 4 questions, total ways = 6⁴ = 1296. This includes the one case where all questions are left unattempted, which must be excluded. Required ways = 1296 − 1 = 1295. Option A (1296) forgets to remove the 'nothing attempted' case, and Option E (624) is 5⁴ − 1, which ignores the option of leaving a question blank. Hence, option D.Q12. In a league, every team plays every other team twice, once at home and once away. If a total of 156 matches were played, how many teams were in the league?Correct Option C … Explanation: Each pair of teams plays twice, so the total number of matches = 2 × nC2 = n(n − 1). Setting n(n − 1) = 156 gives n = 13, since 13 × 12 = 156. Option A (12) gives 132 matches and Option B (14) gives 182 matches. Option D (26) is double the correct value. Hence, option C.Q13. Six fair coins are tossed simultaneously. What is the probability of getting exactly 4 heads?Correct Option B … Explanation: Total outcomes when 6 coins are tossed = 2⁶ = 64. Ways to get exactly 4 heads = 6C4 = 15. Required probability = 15/64. Option A (5/16 = 20/64) corresponds to exactly 3 heads (6C3 = 20), not 4. Hence, option B.Q14. Two fair dice are rolled. The probability of getting a sum of x equals the probability of getting a sum of (x + 4). What is the probability of getting a sum of (x − 2)?Correct Option D … Explanation: The distribution of sums of two dice is symmetric about 7, so P(sum = k) = P(sum = 14 − k). For P(x) = P(x + 4), we need x + (x + 4) = 14, which gives x = 5. Then x − 2 = 3. A sum of 3 occurs as (1, 2) or (2, 1), i.e. 2 outcomes out of 36. Required probability = 2/36 = 1/18. Option B (1/12) is P(sum = 4), and Option A (1/9) is P(sum = 5), i.e. P(x) itself. Hence, option D.Q15. A box contains 5 white, 4 black and 6 blue balls. Two balls are drawn at random. What is the probability that both balls are of the same colour?Correct Option E … Explanation: Total balls = 15, so total ways to draw 2 balls = 15C2 = 105. Both white: 5C2 = 10. Both black: 4C2 = 6. Both blue: 6C2 = 15. Favourable outcomes = 10 + 6 + 15 = 31. Required probability = 31/105. The other options come from miscounting one of the colour cases. Hence, option E.Q16. A bag contains 7 red, 5 green and 3 blue balls. Three balls are drawn at random. What is the probability that at least one ball is green?Correct Option A … Explanation: Total balls = 15, so total ways = 15C3 = 455. Ways with no green ball (drawing from the 10 non-green balls) = 10C3 = 120. P(no green) = 120/455 = 24/91. P(at least one green) = 1 − 24/91 = 67/91. Option B (24/91) is the probability of drawing NO green ball. Hence, option A.Q17. The probability that a shooter hits a target in a single attempt is 0.2. What is the probability that the target is hit at least once in 4 independent attempts?Correct Option C … Explanation: The probability of missing in one attempt = 1 − 0.2 = 0.8. The probability of missing in all 4 attempts = 0.8⁴ = 0.4096. P(at least one hit) = 1 − 0.4096 = 0.5904. Option A (0.4096) is the probability of missing every time, and Option D (0.8) wrongly adds 0.2 four times. Hence, option C.Q18. Two students, A and B, attempt a problem independently. The probability that A solves it is 2/3 and that B solves it is 3/5. What is the probability that exactly one of them solves the problem?Correct Option D … Explanation: P(A solves, B fails) = 2/3 × 2/5 = 4/15. P(A fails, B solves) = 1/3 × 3/5 = 3/15. P(exactly one solves) = 4/15 + 3/15 = 7/15. Option A (2/5 = 6/15) is the probability that BOTH solve it. Hence, option D.Q19. Three archers A, B and C hit a target with probabilities 1/2, 1/3 and 1/4, respectively. If each shoots once, what is the probability that at least two of them hit the target?Correct Option C … Explanation: Exactly two hit: A and B only = 1/2 × 1/3 × 3/4 = 3/24; A and C only = 1/2 × 2/3 × 1/4 = 2/24; B and C only = 1/2 × 1/3 × 1/4 = 1/24. All three hit = 1/2 × 1/3 × 1/4 = 1/24. P(at least two hit) = 3/24 + 2/24 + 1/24 + 1/24 = 7/24. Option A (5/24) is the probability that exactly two hit, missing the case where all three hit. Hence, option C.Q20. What is the probability that in a random arrangement of the letters of the word 'FRAGILE', all the vowels come together?Correct Option B … Explanation: 'FRAGILE' has 7 distinct letters, so total arrangements = 7! = 5040. The vowels are A, I and E. Treat them as one unit: 5 units can be arranged in 5! = 120 ways, and the vowels within the unit in 3! = 6 ways, giving 720 favourable arrangements. Required probability = 720/5040 = 1/7. Option A (1/6) and the other options result from miscounting the units or ignoring the internal arrangement of the vowels. Hence, option B.
