InterviewSolution
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InterviewSolution
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Two numbers a and b are chosen at random from the set {1,2,3,..,5n}. The probability that `a^(4)-b^(4)` is divisible by 5, isA. `(17 n-5)/(25n-1)`B. `(17n+5)/(5(5n-1))`C. `(17n-5)/(5(5n-1))`D. none of these |
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Answer» Correct Answer - C The number of ways of choosing a and b from the given set of 5n integers is `.^(5n)C_(2)`. Let us divide the given set of 5n integers in 5 groups as follows : `G_(1) : 1,6,11, ..,5n-4` `G_(2) : 2, 7,12,.., 5n-3` `G_(3) : 3, 8, 13, ..,5n-2` `G_(4) : 4, 9, 14, .., 5n-1` `G_(5) : 5,10, 15, .., 5n` We have, `a^(4)-b^(4)=(a-b)(a^(2)+b^(2))` So, we observe that `a^(4)-b^(4)` will be divisible by 5, if both a and b belong to the last group or if they belong to any of the remaining four groups. Thus, the number of favourable elementary events is `.^(n)C_(2)+ .^(4n)C_(2)`. Hence, required probability `=(.^(n)C_(2)+.^(4n)C_(2))/(.^(5nC_(2))=(17n-5)/(5(5n-1))` |
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