Find the coefficient of $(a^{5} + b^{4} + c^{7})$ in the expansion of $(b c + c a + a b)^{8}$
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Solution

$(b c + c a + a b)^{8} = \sum_{r_{1} + r_{2} + r_{3}} \frac{8!}{r_{1}! r_{2}! r_{3}!} \times (b c)^{r_{1}} (c a)^{r_{2}} (a b)^{r_{3}}$

We want the coefficient of $a^{5} b^{4} c^{7}$
Let $\frac{8!}{r_{1}! r_{2}! r_{3}!} = k$
$\Rightarrow (b c + c a + a b)^{8} = \sum k \times a^{r_{2} + r_{3}} b^{r_{1} + r_{3}} c^{r_{1} + r_{2}}$
$r_{2} + r_{3} = 5 (p o w e r o f a = 5) - (1)$
$r_{1} + r_{3} = 4 - (2)$
$r_{1} + r_{2} = 7 - (3)$
$(3) - (2) \Rightarrow r_{2} - r_{3} = 3 - (4)$
$(1) + (4) \Rightarrow 2 r_{2} = 8$
$r_{2} = 4$
$\Rightarrow r_{1} = 3 a n d r_{3} = 1$
So the coefficient $k = \frac{8!}{4! \times 3! \times 1!}$

$= \frac{8 \times 7 \times 6 \times 5}{3 \times 2}$

= 280

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