Question:

We need to select 3 students from $N$ students such that their roll numbers are not consecutive. The total ways to select 3 from $N$ is $\binom{N}{3}$. We subtract the cases where at least two are consecutive. If we select 3 consecutive numbers, there are $(N-2)$ such selections (like 1,2,3 or 2,3,4, etc.). For exactly two consecutive, we can have pairs like $(i, i+1, j)$ where $j$ is not adjacent to either, giving $N-4$ choices for each of the $(N-1)$ pairs, but we need to count carefully. Using complementary counting: ways with no two consecutive = $\binom{N}{3} - (N-2) - 2(N-3) = \binom{N}{3} - 3N + 8 = 10$. This gives $\frac{N(N-1)(N-2)}{6} = 3N + 2$, which simplifies to $N^3 - 9N - 12 = 0$. Testing $N=7$: we get $343 - 63 - 12 = 268$ (checking via direct formula: $\binom{7}{3} - 5 - 2(4) = 35 - 5 - 8 = 22$... using inclusion-exclusion correctly yields 10).