We need the number of ordered pairs $(x,y)$ such that:

• $1 \le x \le 25$
• $1 \le y \le 25$
• $x \ne y$
• $x+y$ is divisible by $5$

Classify the integers from $1$ to $25$ according to their remainder modulo $5$.

The possible residues are $0,1,2,3,4$.

Since $25$ numbers are distributed equally among $5$ residue classes, each class has exactly $5$ numbers.

So:

• remainder $0$: $5$ numbers
• remainder $1$: $5$ numbers
• remainder $2$: $5$ numbers
• remainder $3$: $5$ numbers
• remainder $4$: $5$ numbers

For $x+y$ to be divisible by $5$, the residue pairs must be:

Count each case.

Case 1: $(0,0)$

Both $x$ and $y$ are from the $5$ multiples of $5$. Since $x \ne y$,

Case 2: $(1,4)$

Case 3: $(2,3)$

Case 4: $(3,2)$

Case 5: $(4,1)$

Total number of ordered pairs:

Therefore, the required number of ways is

Comparison with stored correct answer:

Stored correct answer = $120$.

Our derived answer is also $120$, so they agree.