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  4. The number of positive integral solutions of the inequality 3 x+y+z ≤ 30, is

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Q. The number of positive integral solutions of the inequality $3 x+y+z \leq 30$, is

Permutations and Combinations

Solution:

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