Question

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

• A. 7
• B. 6
• C. 8
• D. 4

Answer 1

The correct option is A 7

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
$x_1+1+x_2+2+...+x_5+5=20\Rightarrow x_1+x_2+x_3+x_4+x_5=20-15=5Now,x_1\le x_2\le x_3\le x_4\le x_5$
$\begin{array}{ccccc}{x}_1& x_2& x_3& x_4& x_5\\ 0& 0& 0& 0& 5\\ 0& 0& 0& 1& 4\\ 0& 0& 0& 2& 3\\ 0& 0& 1& 1& 3\\ 0& 0& 1& 2& 2\\ 0& 1& 1& 1& 2\\ 1& 1& 1& 1& 1\end{array}$
So, 7 possible cases will be there.