0

Consider the expansion of $(a+b)^9$. The coefficient of the term $a^7b^2$ is the number of ways to select $7$ a's from the $9$ factors of $a+b$. Why is this the case? Furthermore, this seems like an intuitive proof of why binomial coefficients are related to combinatorics. However is there a rigorous mathematical proof that shows the coefficients are the same as combinations?

James Chadwick
  • 743
  • $(a+b)^9$ is none other than the product of $9$ of $(a+b)$s. If you multiply $7$ $a$s out of these terms and $2$ $b$s out of the rest, you will get the term $a^{7}b^{2}$ – acat3 Jul 17 '23 at 10:27
  • You might want to look into generating functions, although not directly related. – Shivansh Jaiswal Jul 17 '23 at 11:22

1 Answers1

1

Let's consider the expansion $(a+b)^n$. $$(a+b)^n = (a+b)(a+b)...(a+b)$$

In the above product we multiply $(a+b)$ $n$ times, so how many terms will we have after we expand?

At first we calculate $(a+b)(a+b)$ which will be a quantity having $2 \cdot 2 =4$ terms being summed. We then calculate that quantity multiplied by $(a + b)$, so we will have a quantity consisting of $2 \cdot 4 = 8$ terms being summed. I think that it is now clear that after $n$ multiplications we will have a sum of $2^n$ terms.

The fact that we are summing $2^n$ terms is the first hint towards the fact that the coefficients have something to do with the number of ways of choosing a subset of size $k$ from a set of size $n$, since $\Sigma_{k=0}^{n}$ $n \choose k$ = $2^n$.

Now the quantity $a^kb^{n-k}$ will certainly appear in the final sum for any $0\leq k \leq n$ but how many times will it appear?

Well the number of times that $a^kb^{n-k}$ appears is equivalent to the number of ways of choosing which $a′s$ of k brackets to multiply out of the $n$ brackets and then multiplying them by the rest of the $b′s$ in the rest of the $n-k$ brackets without the order mattering which is $n \choose k$

So $(a+b)^n = \Sigma_{k=0}^{n}$ $n \choose k$ $a^kb^{n-k}$

Ghanem
  • 33