Find the coefficient of abcd in the expansion of $(a + b + c + d)^{4}$

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Solution

$(a + b + c + d)^{4} = \sum_{r_{1} + r_{2} + r_{3} + r_{4} = 4} \frac{4!}{r_{1}! r_{2}! r_{3}! r_{4}!} a^{r_{1}} . b^{r_{2}} . c^{r_{3}} . d^{r_{4}}$
We want to find the coefficient of a'b'c'd' comparing the powers with the powers in (1), we get $r_{1} = 1, r_{2} = 1, r_{3} = 1 a n d r_{4} = 1$
$\Rightarrow$ coefficient of abcd $\Rightarrow \frac{4!}{1! \times 1! \times 1! \times 1}$
=4!
=24

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