Question

The number of positive integral solutions of the inequality $3x+y+z\le 30$ is

• A. $1115$
• B. $1215$
• C. $1315$
• D. None of these

Answer 1

The correct option is B $1215$

Let w be the non-negative integer such that $3x+y+z+w=30$

Let $a=x-1,b=y-1,c=z-1,d=w$ then

$3a+b+c+d=25$ here $a,b,c,d\ge 0$...............1

Clearly $0\le a\le 8$, if $a=k$ then $b+c+d=25-3k$................2

No. of non-negative integral solutions of $2$ are

${{}^{n+r-1}C}_r={{}^{3+25-3k-1}C}_{25-3k}={{}^{27-3k}C}_2=\frac{3}{2}(3k^2-53k-234)$

Thus required number is

$=\frac{3}{2}\sum _{k=0}^8(3k^2-53k-234)=\frac{3}{2}[3\times 8\times 9\times \frac{17}{6}-53\times 8\times \frac{9}{2}+234\times 9]=1215$