Question

For the inequation $x_1+x_2+x_3\le 13,$ which of the following is (are) CORRECT ?

• A. The number of non-negative integral solutions is $560$
• B. The number of positive integral solutions is $286$
• C. The number of non-negative integral solutions, when $x_2\ge 3$ is $286$
• D. The number of non-negative integral solutions, when $x_1\ge 2,x_2\ge 4$ is $120$

Answer 1

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

The number of non-negative integral solutions, when $x_1\ge 2,x_2\ge 4$ is $120$

$x_1+x_2+x_3\le 13$
Let $x_4\ge 0$ be a dummy variable so that,
$x_1+x_2+x_3+x_4=13\cdots \left(i\right)$
Number of non-negative integral solutions $={}^{13+4-1}{C}_{4-1}={}^{16}{C}_3=560$

For positive integral solutions
$x_1\ge 1,x_2\ge 1,x_3\ge 1$
Let $x_1=X_1+1,x_2=X_2+1,x_3=X_3+1$
From $\left(i\right),$
$X_1+X_2+X_3+x_4=10,$ where $X_1,X_2,X_3\ge 0$
Number of positive integral solutions
$={}^{10+4-1}{C}_{4-1}={}^{13}{C}_3=286$

Now, when $x_2\ge 3$
let $x_2=X_2+3$
$x_1+x_2+x_3+x_4=13$
$x_1+X_2+3+x_3+x_4=13,$ where $X_2\ge 0$
$\Rightarrow x_1+X_2+x_3+x_4=10$
Number of non-negative integral solutions $={}^{10+4-1}{C}_{4-1}={}^{13}{C}_3=286$

When $x_1\ge 2,x_2\ge 4$
let $x_1=X_1+2,x_2=X_2+4$
$x_1+x_2+x_3+x_4=13$
$\Rightarrow X_1+2+X_2+4+x_3+x_4=13,$ where $X_1,X_2\ge 0$
$\Rightarrow X_1+X_2+x_3+x_4=7$
Number of non-negative integral solutions
$={}^{7+4-1}{C}_{4-1}={}^{10}{C}_3=120$