Permutation & Combination — Set #1 | Practice

{"setId":"59134f6f-7888-4641-bc56-e994a46f4b0c","setTitle":"Permutation and Combination – UPSC CSAT","topicName":"Permutation & Combination","topicSlug":"permutation-combination","setNumber":1,"totalQuestions":10,"questions":[{"id":"72753b93-4623-409d-aedb-1ec59709352f","order":1,"statement":"In an examination, a candidate is required to pass all six different subjects. The number of ways he can fail is:","options":{"A":"63","B":"62","C":"51","D":"57"},"correctOption":"A","solution":"Candidates can fail in 1 or 2 or 3 or 4 or 5 or 6 subjects. Required number of ways = 6C1 + 6C2 + 6C3 + 6C4 + 6C5 + 6C6 = 63. Hence, option (a) is correct."},{"id":"a7961fc7-ee07-4dc4-9089-613c0f56eb7b","order":2,"statement":"Which of the following pair of symbols will define the relation between Quantity I and Quantity II? A. Quantity I = Quantity II B. Quantity I > Quantity II C. Quantity I < Quantity II D. Relation cannot be established. Quantity-I: Find the number of factors of 96. Quantity-II: In how many ways can 4 oranges be picked from a bucket of 6 fruits?","options":{"A":"A","B":"B","C":"C","D":"D"},"correctOption":"C","solution":"Number of factors of a number in form of a^x * b^y * c^z = (x+1)(y+1)(z+1). Since 96 = 2^5 * 3^1, number of factors = (5+1)(1+1) = 12. So Quantity I = 12. Quantity II: Number of ways to pick r items from n items = nCr. Required = 6C4 = 15. So Quantity II = 15. Quantity I < Quantity II. Hence, option (c) is correct."},{"id":"be0893e5-d376-4f96-b6ac-487c4da02e70","order":3,"statement":"A committee of five members is to be formed such that at least 2 men are there on the committee. The members of the committee are chosen from the group of 5 men and 4 women. Find the total number of possible ways in which the committee can be formed.","options":{"A":"90","B":"115","C":"121","D":"150"},"correctOption":"C","solution":"Four possibilities: I. 2 men and 3 women, II. 3 men and 2 women, III. 4 men and 1 woman, IV. 5 men and no woman. Number of ways = (5C2 x 4C3) + (5C3 x 4C2) + (5C4 x 4C1) + 5C5 = (10x4) + (10x6) + (5x4) + 1 = 121. Hence, option (c) is correct."},{"id":"abbc354b-2feb-49a3-aed3-5ecab633fd34","order":4,"statement":"In a cricket championship, there are 36 matches. If each team plays one match with every other team, the number of teams is:","options":{"A":"9","B":"8","C":"7","D":"None of these"},"correctOption":"A","solution":"Let n be the number of teams. nC2 = 36, so n(n-1)/2 = 36, n(n-1) = 72. n^2 - n - 72 = 0, (n-9)(n+8) = 0, so n = 9 (since n cannot be -8). Hence, option (a) is correct."},{"id":"a3ef7af1-51da-4824-8fb9-fdc9d30c511d","order":5,"statement":"A box contains four black balls, four white balls and three red balls. The number of ways in which four balls can be drawn from the box so that at least one of the balls is white is?","options":{"A":"295","B":"272","C":"266","D":"168"},"correctOption":"A","solution":"Given: 4 White, 4 Black, 3 Red balls, need to draw four balls with at least one white ball. Possibility-1: 1 white + 3 others = 4C1 x 7C3 = 4x35 = 140. Possibility-2: 2 white + 2 others = 4C2 x 7C2 = 6x21 = 126. Possibility-3: 3 white + 1 other = 4C3 x 7C1 = 4x7 = 28. Possibility-4: all 4 white = 4C4 = 1. Total = 140+126+28+1 = 295. Hence, option (a) is correct."},{"id":"026d2b30-412a-440c-ad6f-a0dddd43a0ad","order":6,"statement":"A box contains 5 black balls and 7 white balls. In how many ways, 6 balls should be selected such that there are at least 2 balls of each colour?","options":{"A":"920","B":"700","C":"765","D":"805"},"correctOption":"D","solution":"Total black = 5, white = 7. Case I: 2 black and 4 white = 5C2 x 7C4 = 350. Case II: 3 black and 3 white = 5C3 x 7C3 = 350. Case III: 4 black and 2 white = 5C4 x 7C2 = 105. Total = 350+350+105 = 805 ways. Hence, option (d) is correct."},{"id":"b4421ffe-95af-4c2c-b55b-57e97875be97","order":7,"statement":"Find the number of triangles obtained by joining 10 points in a plane. (No three points are colinear)","options":{"A":"120","B":"136","C":"145","D":"125"},"correctOption":"A","solution":"As no three points are collinear, each triplet of points yields a triangle. Ten points give 10C3 = (10x9x8)/(3x2x1) = 120 triangles. Hence, option (a) is correct."},{"id":"d2a03108-d3f7-49b2-a6a8-2831bc33b32d","order":8,"statement":"How many chords can be drawn through 31 points on a circle?","options":{"A":"465","B":"595","C":"515","D":"450"},"correctOption":"A","solution":"A chord requires any two points on the circle. So 31 points yield 31C2 chords. Number of chords = 31C2 = (31x30)/2 = 465. Hence, option (a) is correct."},{"id":"98e91a3e-d85a-4a1f-88a9-e303d7f2c51a","order":9,"statement":"From a bunch of 5 blue pens and 7 red pens, how many selections of 5 pens can be made such that at least 3 of them are blue pens?","options":{"A":"230","B":"239","C":"246","D":"312"},"correctOption":"C","solution":"Possible combinations: I. 3 blue + 2 red, II. 4 blue + 1 red, III. 5 blue + 0 red. = (5C3 x 7C2) + (5C4 x 7C1) + (5C5 x 7C0) = (10x21) + (5x7) + (1x1) = 210+35+1 = 246. Hence, option (c) is correct."},{"id":"12da7d2a-0aae-4854-9d3f-60db8c70e0f1","order":10,"statement":"In how many ways can a man invite 6 friends to a party so that 2 or more of them remain present?","options":{"A":"120","B":"45","C":"57","D":"47"},"correctOption":"C","solution":"Two or more friends can remain present in = 6C2 + 6C3 + 6C4 + 6C5 + 6C6 ways = 15+20+15+6+1 = 57 ways. Hence, option (c) is correct."}]}