Question

The number of non-negative integral solutions to the system of equations $x+y+z+u+t=20$ and $x+y+z=5.$

• A. $336$
• B. $246$
• C. $346$
• D. none of these

Answer 1

The correct option is A

$336$

Use the concepts of non-negative integral solution of an equation

We have to find the number of non-negative integral solutions to the system of equations $x+y+z+u+t=20$ and $x+y+z=5.$

Given $x+y+z+u+t=20..\left(i\right)$

$x+y+z=5..\left(ii\right)$

Put $\left(ii\right)$ in$\left(i\right)$

$\begin{array}{rcl}5+u+t& =& 20\\ u+t& =& 15\dots \left(iii\right)\end{array}$

We know that the number of non-negative integral of an inequation having $n$ variable i.e, $x_1+x_2+x_3+.......+x_n=c$ is given by ${}^{c+n-1}{C}_{n-1}$

Number of non-negative integral solutions of (ii) are $={}^{5+3-1}{C}_{3-1}$

$$\begin{gathered} {=}^7{C}_2 \\ =\frac{7!}{2!\left(7-2\right)!} \\ =\frac{7!}{2!5!} \\ =\frac{7\times 6\times 5!}{2!5!} \\ =21 \end{gathered}$$

Number of non-negative integral solutions of (iii) are$={}^{15+2-1}{C}_{2-1}$

$$\begin{gathered} {=}^{16}{C}_1 \\ =\frac{16!}{1!\left(16-1\right)!} \\ =\frac{16\times 15!}{1!\times 15!} \\ =16 \end{gathered}$$

Total number of solutions $=21\times 16=336$

Hence option A is the correct answer.