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

Answer: 7

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,$x_1\le x_2\le x_3\le x_4\le x_5$
$\begin{array}{ccccc}& & & & \\ x_1& x_2& x_3& x_4& x_5\\ 0& 0& 0& 0& 5\\ 0& 0& 0& 1& 4\\ 0& 0& 0& 2& 3\\ 0& 0& 1& 1& 3\\ 0& 0& 1& 2& 2\\ 0& 1& 1& 1& 2\\ 1& 1& 1& 1& 1\end{array}$
So, 7 possible cases will be there.