Question

The number of non-negative integral solutions of $x_1+x_2+x_3+x_4+x_5\le$n (where n is a nonnegative integer) is

• A. ${}^{(n+3)}{C}_3$
• B. ${}^{(n+5)}{C}_5$
• C. ${}^{(n+4)}{C}_4$
• D. ${}^{(n+6)}{C}_6$

Answer 1

Answer: (B)

${}^{(n+5)}{C}_5$

To find the number of solutions for the inequation $x_1+x_2+x_3+x_4+x_5\le n$ is same as sum of the number of solutions of the equations $x_1+x_2+x_3+x_4+x_5=k$, where $0\le k\le n$.
therefore for the given k, such that $0\le k\le n$., the number of solutions of the equation are ${}^{k+5-1}{C}_{5-1}={}^{k+4}{C}_4$.
Hence the total number of solutions are $\sum _{k=0}^n{}^{k+4}{C}_4={}^4{C}_4+{}^5{C}_4+{}^6{C}_4+\cdots +{}^{n-1+4}{C}_4+{}^{n+4}{C}_4$
Now replace ${}^4{C}_4$ by ${}^5{C}_5$ and use the pascal identity for the first two terms $\sum _{k=0}^n{}^{k+4}{C}_4={}^6{C}_5+{}^6{C}_4+\cdots +{}^{n-1+4}{C}_4+{}^{n+4}{C}_4$.
Continuing the process we get the sum as ${}^{n+5}{C}_5$.