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

$\therefore n_1,n_2,n_3,n_{4}and{n}_5$are positive integers such that $n_1<n_2<n_3<n_4<n_5$
then for $n_1+n_2+n_3+n_4+n_5=20$
If $n_1,n_2,n_3,n_4,$ take minimum values 1,2,3,4 respectively then $n_5=10$, there is only one soution $n_1=1,n_2=2,n_3=3,n_4=5$ i.e., one solution Corresponding to $n_5=9,$we can have, only soution $n_1=1,n_2=n_3=3,n_4=4.$
Corresponding to $n_5=8,$ we can have, only solution
$n_1=1,n_2=2,n_3=3,n_4=6$
or $n_1=1,n_2=2,n_3=4,n_4,=4$
i.e., two solutions.
For $n_5=7,$ we can have $n_1=1,n_2=1n_3=4,n_4=5$ i.e. 2 solutions
For $n_5=6,$ we can have
$n_1=2,n_2=3,n_3=4,n_4=5$
i.e., one solution
Thus there can be 7 solutions.