If two numbers x,y are selected at random from the set {1, 2, 3... 65} then the probability that x⁴–y⁴ is divisible by 5 is

S= {1, 2, 3,………., 65} = $\mathrm{A}\cup \mathrm{B}\cup \mathrm{C}\cup \mathrm{D}\cup \mathrm{E}$
Where A= { 5k/k=1,2, ……,13},
B= {5k+1/k=0, 1, 2 , ………..,12}
C= {5k+2/k=0, 1, 2 , ………..,12},
D= {5k+3/k=0, 1, 2 , ………..,12} and
E = {5k+4/k=0, 1, 2 , ………..,12}
n(A)=n(B)=n(C)=n(D)=n(E)=13 x, y $\in$S,(x $\ne$ y) and
x⁴ y⁴ (x² +y² )(x +y)(x -y)  is divisible by 5 if i) both x, y $\in$ same set A or B or C or D or E ( $\therefore$x- y is divisible by 5) or
ii) x,y respectively belongs to (B,E), (C,D), (B,C),
(B,D), (D,E) or in reverse order[($\therefore$ x + y or x² + y² is divisible by 5] Required probability
$=\left[\frac{\left(5{\times }^{13}{\mathrm{C}}_2\right)+\left(6{\times }^{13}{\mathrm{C}}_1{\times }^{13}{\mathrm{C}}_1\times 2\right)}{\left({ }^{65}{\mathrm{C}}_2\right)}\right]=\frac{27}{40}$