Question

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

• A. $5$
• B. $6$
• C. $7$
• D. $8$

Answer 1

The correct option is

C

$7$

Given,

$n_1<n_2<n_3<n_4<n_5$

$n_1+n_2+n_3+n_4+n_5=20$

Now, we will solve this by using combination method.

When we arrange $n_5$ carry forward moves from 0 to 4, the arrangement looks like-

$4_{(}{C}_0)+\frac{4_{(}{c}_1)}{4}+\frac{4_{(}{C}_2)}{3}+\frac{4_{(}{C}_3)}{2}+\frac{\left(4_{(}{C}_4\right))}{1}$=$[\because n_{(}{C}_r)=n!/r!(n-r)!]$
$=1+1+2+2+1$
$=7$

So, the number of distinct arrangement are seven.

Hence, the correct answer is option (C).