Question

Let $n_1<n_2<n_3<n_4<n_5$ be positive integers such that $n_1+n_2+n_3+n_4+n_5=20$. Then the number of such distinct arrangements $(n_1,n_2,n_3,n_4,n_5)$ is .....................

Answer 1

When $n_5$ takes value from $10$ to $6$ the carry forward moves from $0$ to $4$ which can be arranged in
${}^4{C}_0+\frac{{}^4{C}_1}{4}+\frac{{}^4{C}_2}{3}+\frac{{}^4{C}_3}{2}+\frac{{}^4{C}_4}{1}=7$
Alternate solution
Possible solutions are
$1,2,3,4,10$
$1,2,3,5,9$
$1,2,3,6,8$
$1,2,4,5,8$
$1,2,4,6,7$
$1,3,4,5,7$
$2,3,4,5,6$
Hence, there are $7$ solutions.