Permutation & Combination — Set #2 | Practice

{"setId":"c910da43-66ef-4467-8657-8d6c35941e8f","setTitle":"Permutation and Combination 2 – UPSC CSAT","topicName":"Permutation & Combination","topicSlug":"permutation-combination","setNumber":2,"totalQuestions":10,"questions":[{"id":"0fb019be-c364-4f3a-a6b5-3fafd5765740","order":1,"statement":"Amol want to select 3 books from Library among 9 books of his interest which are there in the library. Of these 9 books, he does not want to borrow Biology Part II, unless Biology Part I is also selected. In how many ways can he choose the three books?","options":{"A":"24","B":"63","C":"49","D":"25"},"correctOption":"B","solution":"There are 3 cases: (i) He borrows Biology I, Biology II and any one of remaining 7 books - 7 ways. (ii) He doesn't borrow any Biology book, i.e. borrows 3 books from remaining 7 - 7C3 ways. (iii) He borrows Biology I but not Biology II, so 2 books from remaining 7 - 7C2 ways. Total = 7 + 7C3 + 7C2 = 7 + 35 + 21 = 63. Hence, option (b) is correct."},{"id":"81250dd1-d291-47a2-b0e0-2180b64d2be7","order":2,"statement":"Find the number of different sums that can be formed with the 1 coin of each denomination of Re. 1, Rs. 2, Rs. 5, and Rs. 10 taking at least two denominations at a time.","options":{"A":"15","B":"12","C":"14","D":"11"},"correctOption":"D","solution":"Number of denominations = 4. Case 1: all 4 denominations = 4C4 = 1. Case 2: 3 denominations = 4C3 = 4. Case 3: 2 denominations = 4C2 = 6. Total number of sums = 1+4+6 = 11. Hence, option (d) is correct."},{"id":"c7040c00-d448-4b4b-b1d6-2d5512ea8954","order":3,"statement":"A bag contains 7-black balls, 5-pink balls and 3-white balls. What is the total number of ways in which 4 balls can be withdrawn from the bag such that at least one of the balls is pink?","options":{"A":"1221","B":"1155","C":"987","D":"1320"},"correctOption":"B","solution":"Possibility-1: 1 pink + 3 others = 5C1 x 10C3 = 5x120 = 600. Possibility-2: 2 pink + 2 others = 5C2 x 10C2 = 10x45 = 450. Possibility-3: 3 pink + 1 other = 5C3 x 10C1 = 10x10 = 100. Possibility-4: all 4 pink = 5C4 = 5. Total = 600+450+100+5 = 1155. Hence, option (b) is correct."},{"id":"0d11028c-1835-4395-825e-4a7ff1d260fa","order":4,"statement":"In the question, two Quantities I and II are given. You have to solve both the Quantity to establish the correct relation between Quantity-I and Quantity-II and choose the correct option. Quantity-I: Find the number of unique sets of 7 cards that can be picked from a set of 12 different cards. Quantity-II: A group has 8 men and 6 women. Find the number of committees of 10 members that can be formed from this group such that there are at least 6 men in the group.","options":{"A":"Quantity-I > Quantity-II","B":"Quantity-I < Quantity-II","C":"Quantity-I <= Quantity-II","D":"Quantity-I = Quantity-II or No relation"},"correctOption":"A","solution":"Quantity I: Number of unique sets of 7 cards from 12 = 12C7 = 792. Quantity II: committee with 6 men and 4 women = 8C6 x 6C4 = 28x15 = 420; with 7 men and 3 women = 8C7 x 6C3 = 8x20 = 160; with 8 men and 2 women = 8C8 x 6C2 = 1x15 = 15. Total = 420+160+15 = 595. So Quantity I (792) > Quantity II (595). Hence, option (a) is correct."},{"id":"f50abdaa-57a6-4245-aea9-eaa70c397c18","order":5,"statement":"Find the maximum number of diagonals that can be drawn in Decagon.","options":{"A":"30","B":"35","C":"55","D":"45"},"correctOption":"B","solution":"For an n-sided polygon, nC2 gives total sides plus diagonals. Number of diagonals = n(n-1)/2 - n = n(n-3)/2. For a decagon, n=10, so diagonals = 10(10-3)/2 = 35. Hence, option (b) is correct."},{"id":"642a1b55-f5f8-4da4-8a01-d9320a0177c9","order":6,"statement":"A committee constituting 5 men and 5 women is to be formed. There are a total of 10 women and 8 men. Anita is not ready to be a part of the committee if any of her 3 male friends are a part of the committee. In how many possible ways can the committee be formed?","options":{"A":"7182","B":"7056","C":"1200","D":"1800"},"correctOption":"A","solution":"Number of women = 10, men = 8; need 5 men and 5 women. Case 1: Anita selected - remaining 4 women from 9 and 5 men from remaining 5 (excluding her 3 male friends) = 9C4 x 5C5 = 126x1 = 126. Case 2: Anita not selected - 5 women from 9 and 5 men from 8 = 9C5 x 8C5 = 126x56 = 7056. Total = 7056+126 = 7182. Hence, option (a) is correct."},{"id":"4e60cf83-c2c9-416a-9aa8-0028d7031f5c","order":7,"statement":"A student is required to solve 5 out of the 8 questions in a test. The questions are divided into two sections of 4 questions each. In how many ways can the student select the questions to solve if a minimum of 2 questions must be chosen from either section?","options":{"A":"48","B":"36","C":"12","D":"42"},"correctOption":"A","solution":"Possibility-1: 2 questions from Section 1 and 3 from Section 2 = 4C2 x 4C3 = 6x4 = 24. Possibility-2: 3 questions from Section 1 and 2 from Section 2 = 4C3 x 4C2 = 4x6 = 24. Total = 24+24 = 48. Hence, option (a) is correct."},{"id":"00620b5b-bbc4-4260-a7af-6de5e190b3a6","order":8,"statement":"Find the number of straight lines obtained by joining 10 points on a plane if five of them are collinear.","options":{"A":"40","B":"36","C":"45","D":"55"},"correctOption":"B","solution":"Total lines from 10 points = 10C2 = 45. Since 5 points are collinear, they together form only one line instead of 5C2=10 lines. Required lines = 45 - 10 + 1 = 36. Hence, option (b) is correct."},{"id":"6e5f12b0-8296-406e-86c8-c1d422da0731","order":9,"statement":"A person has to choose 2 drinks out of 5 drinks in a menu of a restaurant, namely, tea, coffee, mango shake, banana shake and orange juice. Orange Juice can be offered only if banana shake is also opted. The number of different combinations of two drinks which can be opted is.","options":{"A":"5","B":"6","C":"7","D":"8"},"correctOption":"C","solution":"If orange juice is not opted, two drinks out of remaining four can be chosen: 4C2 = 6 ways. If orange juice is chosen, it requires banana shake also, giving 1 more way. Total = 6+1 = 7. Hence, option (c) is correct."},{"id":"f9397666-8b30-4164-938e-b0bf29b03752","order":10,"statement":"How many different sums can be formed with the denominations of Rs. 10, Rs. 20, Rs. 50, Rs. 100, and Rs. 500 taking at least three denominations at a time?","options":{"A":"16","B":"17","C":"18","D":"24"},"correctOption":"A","solution":"Number of denominations = 5. Case 1: using all 5 = 1 way. Case 2: using 4 out of 5 = 5C4 = 5 ways. Case 3: using 3 out of 5 = 5C3 = 10 ways. Total = 1+5+10 = 16. Hence, option (a) is correct."}]}