find the coefficient of the given term when the expression is expanded by the binomial theorem

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I am just trying to understand why the term is $\binom{15}8$(3p$^2$ - 2q)$^7$.

I need to find the coefficient in $p^{16}q^7$ in $(3p^2 - 2q)^{15}$

So, I know that $n = 15$ and I have $a^{n - k}b^k$ but I cannot figure out how to get $\binom{15}8$. I've tried watching videos on YouTube and looking up some tutorials but I only found them confusing. I don't have much useful work to add here. The book gives an answer but no explanation and I am completely stuck trying to figure out this.

I am not looking for the complete answer. Just enough to get $\binom{15}8$.

Thanks for any help.

Tony

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You will need the binomial expansion $$(a+b)^{15}=a^{15}+\binom{15}1a^{14}b+\cdots+\binom{15}{k}a^{15-k}b^k+\cdots+b^{15}\ .$$ Substituting $a=3p^2$ and $b=-2q$ gives $$(3p^2-2q)^{15}=\cdots+\binom{15}{k}(3p^2)^{15-k}(-2q)^k+\cdots\ .$$ Now look at the general term $$\binom{15}{k}(3p^2)^{15-k}(-2q)^k$$ and work out what value of $k$ you need in order to get a $p^{16}q^7$ term; the coefficient of this term is $$\binom{15}{k}3^{15-k}(-2)^k$$ and this will be your answer.

To match the given answer you will also need to use the fact that $$\binom{n}{k}=\binom{n}{n-k}\ .$$

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We have the following formula: $$(x+y)^{n}=\sum_{k=0}^n\binom{n}{k}x^ky^{n-k} $$ called the binomial formula. When $k=8$ we have (take $x=3p^2$ and $y=-2q$ in the binomial formula) the term $\binom{15}{8}(3p^2)^{8}(-2q)^{15-8}$.