Question

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

Answer 1

Plan Reducing the equation to a newer equation, where sum of
variables is less. Thus, finding the number of arrangements
becomes easier
As, $n_1\ge 1, n_2 \ge 2,n_3 \ge 3, n_4 \ge 4, n_5 \ge 5$
Let $n_1- 1, = x_1 \ge 0, n_2-2 =x_2 \ge 0, ...,n_5 -5 =x_5 \ge 0$
$\Rightarrow $ New equation will be
$x_1+1+x_2+2+..+x_5=20$
$\Rightarrow \ \ x_1+x_2+x_3+x_4+x_5=20-15=5$
Now,$ \ \ \ x_1 \le x_2 \le x_3 \le x_4 \le x_5 $
$\begin{array}{|c|c|c|} \\ \hline x_1 & x_2 & x_3 & x_4 & x_5 \\ \hline 0&0&0&0&5\\ 0&0&0&1&4\\ 0&0&0&2&3\\ 0&0&1&1&3\\ 0&0&1&2&2\\ 0&1&1&1&2\\ 1&1&1&1&1 \\ \hline \end{array}$
So, 7 possible cases will be there.