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

Answer: (B)

Let $w$ be a non-negative integer such that
$3x+y+z+w=30$
Let $a=x-1,b=y-1,c=z-1,d=w$,
then $3a+b+c+d=25$,
where $a,b,c,d\ge 0(1)$
Clearly, $0\le a\le 8$.
If $a=k$, then
$b+c+d=25-3k(2)$
Number of non-negative integral solutions of equation (2)
$={}^{n+r-1}{C}_r={}^{3+25-3k-1}{C}_{25-3k}$
$={}^{27-3k}{C}_{25-3k}={}^{27-3k}{C}_2$
$=\frac{(27-3k)(26-3k)}{2}$
$=\frac{3}{2}\left(3k^2-53k-234\right)$
$\therefore$ Required number $=\frac{3}{2}\sum _{k=0}^8\left(3k^2-53k+234\right)$
$=\frac{3}{2}\left[3\cdot \frac{8\times 9\times 17}{6}-53\frac{8\times 9}{2}+234\times 9\right]=1215$