Question

The number of non-negative integral solutions of $x_1+x_2+\dots +x_{10}\le 15$ is

• A. ${}^2{}^{5}C{}_1{}_0$
• B. ${}^1{}^{4}C{}_1{}_0$
• C. ${}^2{}^{5}C{}_9$
• D. ${}^2{}^{6}C{}_9$

Answer 1

Answer: (A)

${}^2{}^{5}C{}_1{}_0$

We have $x_1+x_2+...+x_9+x_{10}\le 15$, where $x_{i}s$ are non-negative integers. To find the number of solutions, let us introduce a variable $x_{11}$.
Let $x_1+x_2+x_3+...+x_9+x_{10}+x_{11}=15$ , where $x_{11}$ is also a non-negative integer.
The number of non-negative integral solutions of $x_1+x_2+...+x_r=n$ is ${}^{n+r-1}{C}_{r-1}$.
$n=15;r=11$
Hence, the number of solutions in this case ${=}^{25}{C}_{10}$
This is same as the number of non-integral solutions of the inequality.

Hence, option A is correct.