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, is

Options

A. `(17 n-5)/(25n-1)`
B. `(17n+5)/(5(5n-1))`
C. `(17n-5)/(5(5n-1))`
D. none of these

Answer

C

Explanation

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))`