Mathematics-4.Unit-05-Permutation & Combination Test With Solution

1. (A) A number is divisible by four, if the last two digits are divisible by four. In this case last two digits can be 12, 16, 28, 32, 36, 68, 92, or 96. Total number of such numbers = 8. (4C3 . 3!) = 192. 2. (C) Number of ways = = 12 3. (A) 1M, 4l, 2P and 4S. Total gaps created by letter 1M, 4l and 2P is 8. Thus total arrangements = . 4. (B) Number of words will be equal to the number of ways in which 9 letters, in which 2 alive, 2 other alive and other alive can be arranged = = . 5. (A) Total words, without any restriction = 7! Total words beginning with l = 6! Total words ending with B = 6! Total words with l and ending with B = 5! Thus total number of required words = 7! – (6! + 6! – 5!) = 7! – 2(6!) + 5! 6. (A) Number of numbers which are multiple of 5 = = 25 Number of numbers which are multiple of 25 = = 5 Number of numbers which are multiple of 25 = = 1 So number of zeroes at the end of 127! = number of multiples of line + number of multiples of 25 + number of multiples of 125 = 25 + 5 + 1 = 31 7. (D) A number is divisible by three if the sum of it’s digits is divisible by 3. Sum of the digits of the given set is 15. That means either we leave out zero or 3. Total numbers when zero is left out = 5! = 120 Total numbers when three is left out = 4.4.3.2.1 = 96 Thus number of such numbers = 216. 8. (B)      … N places, for positive sign. Now we have to fill N + 1 places between positive signs and end places with negative signs. This ca be domain N + 1Cn ways such that no negative signs are together. 9. (C) Given word have 2A’s, 2N’s, 1S, 1H, 1U, 1M Total words = 10. (B) Let number of persons = n, nC2 = 66  = 66  n (n  1) = 132  n = 12 11. (B) That means formed number can be almost of three digits. Total number of one digit numbers = 6 Total number of two digit numbers = 5.5 = 25 Total number of three digit numbers = 5.5.4 = 100 Thus total number of such numbers = 131 12. (B) (a2  a)C2 = (a2  a)C4 then a2  a = 4 + 2  a = 3, a =  2 a2  2 is not possible so a  3 13. (D) There will be (n1 + 1) gaps created by n1 men. Now women have to be seated only in these gaps. Thus number of such sitting arrangements = 14. (C) Number of triangles formed = 12C3  7C3 = 185 15. (B) There will be 4 gaps created by 4 boys. Girls can sit only in these gaps. Boys can be seated in 3! Ways and there after girls can be seated in 4! ways. 16. (B) First of all two men can be selected in 9C2 ways. Thereafter 2 women can selected in 7C2 ways (as wives of selected men are not be selected). And finally they can be paired up in 2C1 ways. Thus total ways = 17. (D) nCr3 + nCr2 + 2 (nCr2 + nCr1) + nCr1 + rCr = n+1Cr2 + 2 n+1Cr1 + n+1Cr = n+1Cr2 + n+1Cr1 + n+1Cr1 + n+1Cr = n+2Cr1 + n+2Cr = n+3Cr. 18. (A) It is same as dividing 22 books in 5 packets. Such that two packets have 5 books each and remaining three have four books each and then distributing these packets. Thus total ways of distributing these books = = 19. (D) Total number of triangles = + = 3.4.5. + 3.9 + 6.8 +10.7 = 205 20. (B) The number of matches in the first round = 6C2+6C2. The number of matches in the next round = 6C2 The number of matches in the semifinal round round = 4C2. So, the required number of matches = 6C2+6C2+6C2+4C2 +2 = 53 (Note: For “ best of three” at least two matches are played.) 21. (D) Total permutations = 0 + k + k2 +k3 + ……… + kr = 22. (B) The number of times the teacher goes to the zoo = nC3. The number of times a particular child goes to the zoo = n-1C2. From the question , nC3 – n-1C2 = 84. Or (n -1)(n –2)(n –3) = 6  84 = 9 8  7  n – 1 = 9  n = 10 . 23. (C) Total variables if only the alphabet is used is equal to 26. Total variables if alphabets and digits both are used = 26.10 Total variables = 26(1 + 10) = 286 24. (C) A = {2, 3, 5, 7, 11,13, 17, 19, 23, 29}. A rational number is made by taking any two in any order . So, the required number of rational numbers = 10P2 +1 ( including 1). 25. (A) Let xw, xR, xB be the number of white balls, red balls and blue balls being selected. We must have xW + xR + xB 10. Required ways = number of non-negative integral solutions of xW + xR + xB = 10 = 3 + 10 –1C10 = 12C10 = 12C2 26. (B) Since 3 does not occur in 1000, we have to count the number of times 3 occurs when we list the integers from 1 to 999. Any number between 1 and 999 is of the form xyz where 0  x, y, z 9 . Let us first count the numbers in which 3 occurs exactly once. Since 3 can occur at one place in 3C1 ways, there are 3C1 ( 99) = 3 92 such numbers. Next, 3 can occur exactly two places in (3C2) (9) = 3  9 such numbers. Lastly , 3 can occur in all three digits in one number only. Hence the number of times 3 occurs is 1  ( 3 92) + 2 ( 39) + 31 = 300 27. (A) Matches whose predictions are correct an be selected in 20C10 ways. Now each wrong prediction can be made in 2 ways. Thus total ways = 20C10 .210 28. (C) Let x1 ,x2 , x3, x4 be the number of white, red, blue, green balls respectively that are selected. Then x1 +x2+ x3+x4 = 10. The required of ways = coefficient of y10 in ( 1+ y+ y2 + y3 + . . . )4 = coefficient of y10 in(1 - y)-4 = coefficient of y10 in ( 1+ 4C1 y + 5C2y2 +6C3y3 + . . . = . 29. (C) Let S1 and S2 refuse to be together and S3 and S4 want to be together only. Total ways when S3 and S4 are selected = (8C2 + 2C1 . 8C1) = 44 Total ways when S3 and S4 are not selected = (8C4 + 2C1.8C1) = 182 Thus total ways = 44 + 182 = 226 30. (B) We can choose two men out of 9 in 9C2 ways. Since no husband and wife are to play in the same game, two women out of the remaining 7 can be chosen in 7C2 ways. If M1, M2 , W1 and W2 are chosen, then a team may consist of M1 and W1 or M1 and W2. Thus the number of ways of arranging the game is ( 9C2)( 7C2)(2) = = 1512.