Let $A$ denotes the event that the difference between the first and second number is at least $m.$
Let ${E_x}$ denote the event that the first number chosen is $x,$
we must have $x - y \ge m$ or $y \le x - m.$
Therefore $x > m$ and $y < n - m.$
Thus $P({E_x}) = 0$ for $0 < x \le m$ and $P({E_x}) = \frac{1}{n}$ for $m < x \le n.$
Also $P(A/{E_x}) = \frac{{(x - m)}}{{(n - 1)}}$
Therefore, $P(A) = \sum\limits_{x = 1}^n {P({E_x})\,\,P(A/{E_x})} $
$ = \sum\limits_{x = m + 1}^n {P({E_x})\,\,P(A/{E_x})} = \sum\limits_{x = m + 1}^n {\frac{1}{n}.\frac{{x - m}}{{n - 1}}} $
$ = \frac{1}{{n(n - 1)}}[1 + 2 + 3 + ..... + (n - m)]$
$ = \frac{{(n - m)\,(n - m + 1)}}{{2n(n - 1)}}.$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.