How can I find the coefficient of $a^3b^2$ in the expansion of:
$$ (a+b+3)^5 $$
What is the quickest way to do this without doing the whole expansion?
$\endgroup$ 1Answer:
Coefficient of $a^3b^2$ is $10$ according toWolframAlpha
3 Answers
$\begingroup$You're looking for the multinomial theorem and coefficients.
$$(a+b+3)^5 = \sum_{k_1+k_2+k_3=5} \frac{5!}{k_1!k_2!k_3!}a^{k_1}b^{k_2}3^{k_3}$$
We're looking for $k_1=3, k_2=2, k_3=0$. The corresponding multinomial coefficient is
$$\frac{n!}{k_1!k_2!k_3!} = \frac{5!}{3!2!0!} = \frac{120}{12}=10.$$
Note that $\binom{5}{3}\binom{2}{2} =\frac{5!}{3!2!}\frac{5!}{5!0!} = \frac{5!}{3!2!0!}$ as in uniquesolution's answer.
$\endgroup$ 3 $\begingroup$Write $$(a+b+3)^5 = (a+b+3)(a+b+3)(a+b+3)(a+b+3)(a+b+3)$$ You want to choose $3$ $a$'s and two $b$'s. You have ${5 \choose 3}=10$ ways to choose the $a$'s, and then you must choose $b$'s from the other two parentheses, so you have $10$ ways.
$\endgroup$ $\begingroup$Definitely not the quickest way, but:
$$(a+b+3)^5=((a+b)+3)^5$$
which produces powers of $a+b$, so the multinomial coefficient is the binomial coefficient on the one free power of $a+b$, which is the only one that contains $a^3b^2$, and is $\binom{5}{2}=10$
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