Permutation & Combination — Set #4 | Practice

{"setId":"0c7f5080-8702-4afa-a85a-a3c9c975a04d","setTitle":"Permutation and Combination 4 – UPSC CSAT","topicName":"Permutation & Combination","topicSlug":"permutation-combination","setNumber":4,"totalQuestions":10,"questions":[{"id":"bebae5ef-f8db-44f7-b7cf-fda9b7097f79","order":1,"statement":"A flag with 5 vertical strips has to be designed using some or all of the colours i.e. white, green, red, and blue. Find the number of different ways to design a flag so that no two adjacent stripes have the same colour.","options":{"A":"128","B":"198","C":"324","D":"364"},"correctOption":"C","solution":"For the first stripe, number of ways = 4. For each subsequent stripe (2nd through 5th), the colour must differ from the previous one, giving 3 choices each. Total = 4 x 3 x 3 x 3 x 3 = 324. Hence, option (c) is correct."},{"id":"da816658-51eb-4f31-9480-2053f858a08a","order":2,"statement":"In an examination, there are four subjects each with maximum marks of 100 and marks allotment in each subject are in integers only. In how many different ways a student can score 99% marks?","options":{"A":"20","B":"35","C":"23","D":"30"},"correctOption":"B","solution":"Total marks required = 400 x 99/100 = 396, meaning 4 marks are deducted overall in some combination across the 4 subjects. Case (4,0,0,0): 4!/3! = 4 ways. Case (3,1,0,0): 4!/2! = 12 ways. Case (2,2,0,0): 4!/(2!x2!) = 6 ways. Case (2,1,1,0): 4!/2! = 12 ways. Case (1,1,1,1): 1 way. Total = 4+12+6+12+1 = 35. Hence, option (b) is correct."},{"id":"d85565bf-98f3-41a3-b3f7-f9ec2c5622e8","order":3,"statement":"What are the total number of 3-digit positive integers that have at least one digit repeated?","options":{"A":"252","B":"728","C":"152","D":"304"},"correctOption":"A","solution":"Total 3-digit integers = 900. Number of 3-digit integers with no repeated digits: first digit has 9 options (1-9), second has 9 options (0-9 except first), third has 8 options (excluding first two) = 9x9x8 = 648. Numbers with at least one repeated digit = 900 - 648 = 252. Hence, option (a) is correct."},{"id":"ac6b0380-05d2-4a62-91b2-afa45c5e6b25","order":4,"statement":"When four dice are rolled simultaneously with each face marked 1 to 6, then what are the number of possible ways such that at least two dice show the number 3?","options":{"A":"121","B":"151","C":"171","D":"181"},"correctOption":"C","solution":"Case 1: exactly two dice show 3 = 4C2 x 5 x 5 = 150. Case 2: exactly three dice show 3 = 4C3 x 5 = 20. Case 3: all four dice show 3 = 4C4 = 1. Total = 150+20+1 = 171. Hence, option (c) is correct."},{"id":"3d95a02d-5994-479d-b53e-f325244b262e","order":5,"statement":"In how many ways can one make a flag of three stripes, if there are 5 red stripes, 6 white stripes and 4 black stripes available?","options":{"A":"455","B":"1005","C":"1055","D":"1657"},"correctOption":"D","solution":"Case 1: three stripes of the same colour = 5C3 + 6C3 + 4C3 = 34. Case 2: two stripes of the same colour combined with arrangement, computed separately for red (300), white (405), and black (198) pairs, totaling 903. Case 3: three stripes of different colours arranged = (5C1 x 6C1 x 4C1) x 3! = 720. Total = 34+300+405+198+720 = 1657. Hence, option (d) is correct."},{"id":"4e7c7d7d-251a-47f4-bae5-e0c0e6e5aebf","order":6,"statement":"How many five digit numbers can be formed using the digits 0, 3, 4, 5, 6 and 7 which are divisible by 5 without repetition of the digits?","options":{"A":"210","B":"216","C":"213","D":"225"},"correctOption":"B","solution":"For divisibility by 5, the unit digit must be 0 or 5. Case ending in 0: remaining 4 digits arranged from 5 choices = 5x4x3x2 = 120. Case ending in 5: first digit cannot be 0, giving 4 choices, then remaining two positions have 4 and 3 choices, then last remaining has... total = 4x4x3x2 = 96. Total required numbers = 120+96 = 216. Hence, option (b) is correct."},{"id":"880d7fd6-5920-4a34-a796-49faa3e46fdd","order":7,"statement":"Find the number of ways in which exactly two letters are put into incorrect envelopes, if there are six letters and six envelopes and exactly one letter is to be put in exactly one envelope with the correct address.","options":{"A":"6","B":"9","C":"12","D":"15"},"correctOption":"D","solution":"Exactly two letters out of six go into incorrect envelopes, meaning four letters go into correct envelopes. Number of ways to select and correctly place 4 letters = 6C4. The remaining two letters can be swapped incorrectly in exactly 1 way. Total = 6C4 x 1 = 15. Hence, option (d) is correct."},{"id":"4fe09952-6dcc-4302-9bd6-53de4cf2e2e3","order":8,"statement":"How many 3-digit numbers that are divisible by 4 can be formed using the digits 2, 4, 7 and 8 (repetition of digits is not allowed)?","options":{"A":"14","B":"15","C":"20","D":"10"},"correctOption":"D","solution":"A number is divisible by 4 if its last two digits are divisible by 4. Possible last-two-digit pairs (XY) from the given digits are 24, 28, 48, 72, or 84 - 5 cases. The first (hundreds) position can then be filled by 1 of the 2 remaining digits. Total numbers = 5 x 2 = 10. Hence, option (d) is correct."},{"id":"87727fa7-8cab-4908-a607-adf71859986b","order":9,"statement":"Find the total number of ways to rearrange the word BHUTAN such that the rearranged word starts and ends with a vowel.","options":{"A":"48","B":"12","C":"24","D":"32"},"correctOption":"A","solution":"BHUTAN has 2 vowels (A, U). The rearranged word must start and end with a vowel, which can be arranged in 2 ways (A...U or U...A). The remaining 4 letters can be arranged in the middle in 4! ways. Total = 2 x 4! = 2 x 24 = 48. Hence, option (a) is correct."},{"id":"1fe72d40-e181-4548-8d31-9f6ef236df55","order":10,"statement":"Find the number of different ways in which a batsman can score exactly 21 runs by scoring single runs, three runs and sixes only.","options":{"A":"19","B":"20","C":"21","D":"22"},"correctOption":"B","solution":"Let x, y, z be the number of singles, threes, and sixes, such that x+3y+6z=21 with x,y,z >= 0. For z=0: x+3y=21 gives 8 solutions. For z=1: x+3y=15 gives 6 solutions. For z=2: x+3y=9 gives 4 solutions. For z=3: x+3y=3 gives 2 solutions. Total = 8+6+4+2 = 20. Hence, option (b) is correct."}]}