 | Permutations and Combinations |
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1) Sum of all three digit numbers (no digit being zero) having the property that all the digits are perfect squares is a)3108 b)6216 c) 13986 d)none of these 2)In a certain test there are are n questions. In this test 2k students gave wrong answers to atleast (n-k) questions where k=0,1,2,3,4,.......n.If the total number of wrong answers is 4095,then the value of n is a)11 b)12 c)13 d)15 3) An 8 digit number divisible by 9 is to be formed by using 8 digits 0,1,2,3,4,5,6,7,8,9 without replacement . The number of ways in whivh this can be done is a)9! b)2(7!) c)36(7!) d)4(7!) |
| Re: Permutations and Combinations |
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No.-1 Possible digits: 1, 4, 9 Assumption: Repetition allowed Number of numbers for which 1 is at unit place: 3 x 3 = 9 Number of numbers for which 4 is at unit place: 3 x 3 = 9 Number of numbers for which 9 is at unit place: 3 x 3 = 9 Sum of numbers at unit place: 9x1+9x4+9x9=126 Similarly, sum of numbers at 10th place: 126 Similarly, sum of numbers at 100th place: 126 Total: 1x126+10x126+100x126 |
| Re: Permutations and Combinations |
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Thank you!!! |
| Re: Permutations and Combinations |
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Welcome, it is easier to help if you attach your work. It saves time too. When questions are from world standard texts then also it helps as the question is likely to be well researched and error margin in it is miniscule. No.-2 2k---------> n-k questions, 2n-k ---------> n-(n-k)=k questions at least and 2n-(k+1) ------------->n-n+(k+1)=k+1 questions at least Exactly k questions wrong, 2n-k - 2n-k-1 students Total wrong answers = Sigma k(2n-k - 2n-k-1) where signma is to be taken from k=0 to k=n-1 which comes out to be 2n-1. 2n-1=4095 and n can be known. |
