# Binomial expansion and factorials

I have come across this question, the answer is simply stated as 36, and while I can see how 36 is gotten, I don't understand why? How is it meant to be known that b = 36, just looking at the formula is nice and quick but I much prefer to understand why the statement in the question is true?

Text from the above picture:

Given that $\binom{40}4=\frac{40!}{4!b!}$, write down the value of $b$.

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I've added the text from the picture; see meta. –  Martin Sleziak Jun 22 '12 at 10:05

## 3 Answers

If ${n \choose k}$ is simply notation for $\dfrac{n!}{k!(n-k)!}$ then the answer is immediate.

If ${n \choose k}$ represents the number of ways of choosing $k$ items from $n$ without worrying about order, then it is a combination and it is not difficult to see that this is $\frac{n(n-1)(n-2)\cdots(n-k+1)}{k(k-1)(k-1)\cdots 1}$, which is again $\frac{n!}{k!(n-k)!}$.

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Do you know the following equality? $$\binom{m}{n} = \frac{m!}{n! \; x!}$$

where $x$ is an expression in $m, n.$ What is $x$? If you don't know the answer, then you should read over the binomial coefficient page on Wikipedia.

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The binomial coefficient $\binom{m}{n}$ is defined to be the number of ways of choosing $n$ objects from $m$, with no emphasis on ordering.

Well how many ways are there of doing this? We can chose our first object in $m$ ways, then for each choice we have $m-1$ ways of getting the second object, etc.

So we have $m(m-1)(m-2)...(m-n+1)$ ways of choosing $n$ objects from $m$ WITH ordering considerations. We now divide by the number of ways of reordering $n$ objects.

This is $n(n-1)(n-2)...3.2.1 = n!$

So:

$\binom{m}{n} = \frac{m(m-1)(m-2)...(m-n+1)}{n!} = \frac{m!}{n!(n-m)!}$

(multiplying top and bottom by $(n-m)!$ to tidy things up).

Now it is obvious what the answer to your question is, since here we have $m=40, n=4$, so that $m-n=36$.

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