Question

For the equation $x_1+x_2+x_3\le 13$, the correct option(s) is/are

• A. The number of non negative integral solution will be $560$
• B. The number of positive integral solution will be $220$
• C. The number of non negative integral solution, when $x_2\ge 3$ will be $286$
• D. The number of non negative integral solution, when $x_1\ge 2,x_2\ge 4$ will be $120$

Answer 1

Answer: (A), (B), (C), (D)

The number of non negative integral solution, when $x_1\ge 2,x_2\ge 4$ will be $120$

Here,$x_1+x_2+x_3\le 13$
Let $x_4\ge 0$ be a dummy variable so that, we get
$x_1+x_2+x_3+x_4=13$
Number of non negative integral solution $={}^{13+4-1}{C}_{4-1}={}^{16}{C}_3=560$
Number of positive integral solution $={}^{13-1}{C}_{4-1}={}^{12}{C}_3=220$

Now, when $x_2\ge 3$ let $x_2=X_2+3$
Number of non negative integral solution of
$x_1+x_2+x_3+x_4=13$
$\Rightarrow x_1+X_2+3+x_3+x_4=13$
$\Rightarrow x_1+X_2+x_3+x_4=10$
$={}^{10+4-1}{C}_{4-1}={}^{13}{C}_3=286$

When $x_1\ge 2,x_2\ge 4$ let $x_1=X_1+2,x_2=X_2+4$
Number of non negative integral solution of
$x_1+x_2+x_3+x_4=13$
​​​​​​​$\Rightarrow X_1+2+X_2+4+x_3+x_4=13$
​​​​​​​$\Rightarrow X_1+X_2+x_3+x_4=7$
$={}^{7+4-1}{C}_{4-1}={}^{10}{C}_3=120$