Let `n and k` be positive such that `n leq (k(k+1))/2`.The number of solutions `(x_1, x_2,.....x_k), x_1 leq 1, x_2 leq 2, ........,x_k leq k`, all integers, satisfying `x_1 +x_2+.....+x_k = n`, is .......

Answer

`1/2 (2n - k^(2) + k - 2)`

Explanation

The number of solution of `x_(1) + x_(2) + .. + x_(k) = n`
= Coefficient of `t^(n)` in `(t + t^(2) + t^(3) + ..) (t^(2) + t^(3) + ..) (t^(k) + t^(k + 1) + ..)`
= Coefficient of `t^(n)` in `t^(1 + 2 + .. + k) (1 + t + t^(2) + ..)^(k)`
Now, `1 + 2 + .. + k = (k(k + 1))/(2) = p` [say]
and `1 + t + t^(2) + .. = (1)/(1 - t)`
Thus, the number of required solutions
= Coefficient of `t^(n - p)` in `(1 - t)^(-k)`
= Coefficient of `t^(n - p)` in `[1 + ""^(k)C_(1)t + ""^(k + 1)C_(2)t^(2) + ""^(k + 2)C_(3)t^(3) + ..]`
`= ""^(k + n - p - 1)C_(n - p) = ""^(r)C_(n - p)`
where, `r = k + n - p - 1 = k + n - 1 - 1/2 k(k + 1)`
`= 1/2 (2k + 2n - 2 + k^(2) - k) = 1/2(2n - k^(2) + k - 2)`