Let $x$ and $y$ be distinct integers where $1 \le x \le 25$ and $1 \le y \le 25$. Then, the number of ways of choosing $x$ and $y$, such that $x+y$ is divisible by 5, is ____________.

Let $x+y=5 \lambda$ <br/><br/>Possible cases are <br/><br/>$\begin{array}{llc}x & y & \text { Number of ways } \\ 5 \lambda(5,10,15,20,25) & 5 \lambda(5,10,15,20,25) & 20 \\ 5 \lambda+1(1,6,11,16,21) & 5 \lambda+4(4,9,14,19,24) & 25 \\ 5 \lambda+2(2,7,12,17,22) & 5 \lambda+3(3,8,13,18,23) & 25 \\ 5 \lambda+3(3,8,13,18,23) & 5 \lambda+2(2,7,12,17,22) & 25 \\ 5 \lambda+4(4,9,14,19,24) & 5 \lambda+1(1,6,11,16,21) & 25\end{array}$ <br/><br/>Total number of ways $=20+25+25+25+25=120$ <br/><br/><b>Note :</b> In first case total number of ways = 20 as in the question given that the chosen $x$ and $y$ are distinct integers each time. So when you choose x as 5 then you can't choose y as 5. Possible values of y are (10, 15, 20, 25). So, here four possible pairs of (x, y) possible { (5, 10), (5, 15), (5, 20), (5, 25)}. <br/><br/>Similarly, four possible pairs of (x, y) possible each time when x = 10, 15, 20 and 25. <br/><br/>$\therefore$ Total number of ways in the first case = 5 $\times$ 4 = 20.