Question

Two numbers $x$ and $y$ are chosen at random without replacement from the set $1,2,3,...,5n$.The probability that $x^4-y^4$ is divisible by $5$ is $\frac{(6k-1)n-5}{5(5n-1)}$. Find the value of $k$?

Answer 1

$x^4-y^4$=$(x^2+y^2)(x-y)(x+y)$

Let us consider five sets $(5k,5k+1,5k+2,5k+3,5k+4)$.
Case I: $(x-y)$ is divisible by $5$.
Elements are to be selected from the same set. ${5.}^n{C}_2$ ....(1)

Case II: $x+y$ is divisible by $5$ but $x-y$ is not divisible by $5$
It can be done from $(5k+1,5k+4)$ or $(5k+2,5k+3)$.
Total ways $=2n^2$ ....(2)

Case III: $x^2+y^2$ is divisible by $5$ but $x-y$ or $x+y$ is not divisible by $5$ ...(3)

Type	Remainder of Square
$5k$	$0$
$5k+1$	$1$
$5k+2$	$2$
$5k+3$	$3$
$5k+4$	$4$

We can select one number either from$(5k+1,5k+4)$ and the second number from $(5k+2,5k+3)$.
Hence, the number of cases $=2n\times 2n=4n^2$ ................ (3)

Adding (1), (2), (3)

Total number of favourable cases $=\frac{5.n(n-1)}{2}+6n^2=\frac{17n^2-5n}{2}$

Total number of cases $=\frac{5n(5n-1)}{2}$

Probability $=\frac{17n-5}{25n-5}$

Comparing it with the given expression $k=3$.