by Martynas Krupskis
Last Updated September 11, 2019 18:20 PM - source

How to expand $\dfrac{n!}{k!(n-k)!}$ to $\dfrac{n(n-1)(n-2)\cdots(n-k+1)}{k!}$?

I've seen many proofs of binomial expansion where they assume that both equations are true. But they never explain how they jump to that second form of combinations formula?

What's the way to expand that?

$n!$ is defined as:

$n! = 1 \cdot 2 \cdot 3 \cdots (n-2) \cdot (n-1) \cdot n$ $=\prod_{i = 1}^n i$

Now consider the following example of a simple case

Which can also be written as:

Now come to the general case:

$$(n-0)(n-1)(n-2)\cdots\left(n-(k-1)\right) = \frac{n!}{(n-k)!} = n^{\underline k}$$

Similarly for your equation:

$$\frac{n(n-1)(n-2)\cdots(n-k+1)}{k(k-1)(k-2)\cdots 1}= \frac{n!}{(n-k)!k!} = \binom {n}{k} = \frac{n^{\underline k}}{k!}$$

It's just using the definition of $n!$ and simple division of integers.

September 11, 2019 18:03 PM

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