4
$\begingroup$

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?

$\endgroup$
4
  • 4
    $\begingroup$ Note that:$$\frac{n!}{(n-k)!} = \frac{n (n-1) (n-2) \ldots (n-k+1) \color{blue}{(n-k) (n-k-1) \ldots}}{\color{blue}{(n-k)(n-k-1))\ldots}}$$ $\endgroup$ Sep 11, 2019 at 17:39
  • $\begingroup$ By the way, welcome to Math.SE. Take the opportunity to take the tour, if you haven't done it already. See also some tips on how to ask, on formatting help and on writing down equations using LaTeX / MathJax. $\endgroup$ Sep 11, 2019 at 17:40
  • $\begingroup$ Thanks, but I don't really understand how does n! expand to that. $\endgroup$ Sep 11, 2019 at 17:44
  • $\begingroup$ "I don't really understand how does n! expand to that." To answer that, I remind you that for natural number $n$ you have that $n!$ is defined as $n!=n\times (n-1)\times (n-2)\times \cdots \times 3\times 2\times 1$. You might also see it defined as $n!=\prod\limits_{k=1}^n k$ or defined recursively as $0!=1!=1$ and $n!=n\times (n-1)!$ for each $n>1$. If you instead defined $n!$ as the number of bijective functions from a finite set with $n$ elements to itself, then from there you can show that it is equivalent to the other definitions that I give here. $\endgroup$
    – JMoravitz
    Sep 11, 2019 at 17:58

1 Answer 1

5
$\begingroup$

$n!$ 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

enter image description here

Which can also be written as:

enter image description here

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.

More information 1 2

$\endgroup$
2
  • $\begingroup$ It is worth keeping the reference to $n\frac{k}{~}$ and including a link to the wikipedia page for falling factorials as this is not uncommon notation to encounter. It might not appear in elementary textbooks all of the time, but it is still useful notation and finds its use frequently beyond the elementary courses. $\endgroup$
    – JMoravitz
    Sep 11, 2019 at 18:07
  • $\begingroup$ @JMoravitz I thought he might not be familiar with that notation, so can cause confusion to fairly simple topic. $\endgroup$
    – IamKnull
    Sep 11, 2019 at 18:08

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .