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Prove that $ \frac{n(n-1)(n-k+1)}{k(k-1)\cdots1} = \frac{n!}{(n-k)!k!}$.

$= \frac{n(n-1)...(n-k+1)(n-k)!}{(n-k)!(k(k-1)...1)(3 \times 2 \times 1)}$

$= \frac{(n(n-1)(n-2)...(n-k+1)}{(k(k-1)....1)}$

$= \frac{n(n-1)..(n-k+1)}{k!}$

My question is how is it that $(n-k)!$ is able to appear in the numerator and denominator on the right side of the equation? This is not obvious to me. What is (n-k)! in expanded form?

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    $\frac{n!}{(n-1)!k!}=\frac{n}{k!}$ so something is not written correctly. – 2017-01-21
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    Your title is inconsistent with the body. – 2017-01-21
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    Note : $n(n-1).....(n-k+1) = \frac{n!}{(n-k)!}$ – 2017-01-21
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    I think you meant $\frac{n(n-1)\cdots(n-k+1)}{k(k-1)\cdots1} = \frac{n!}{(n-k)!k!}.$ – 2017-01-21

2 Answers 2

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Hints:

$$n!=1\times2\times3\times\dots\times(n-k-1)\times(n-k)\times(n-k+1)\times\dots\times n$$

$$(n-k)!=1\times2\times3\times\dots\times(n-k)$$

Thus,

$$\begin{align}\frac{n!}{(n-k)!}&=\frac{\color{#4488dd}{1\times2\times3\times\dots(n-k-1)\times(n-k)}\times(n-k+1)\times\dots\times n}{\color{#4488dd}{1\times2\times3\times\dots(n-k-1)\times(n-k)}}\\\vphantom{\cfrac11}&=(n-k+1)\times(n-k+2)\times\dots\times n\end{align}$$

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We have: $$\frac{n!}{(n-k)!}=\frac{n(n-1)(n-2)\ldots(n-k+1)\color{#AA0000}{(n-k)(n-k-1)\ldots\cdot 2 \cdot 1}}{\color{#AA0000}{{(n-k)(n-k-1)\ldots\cdot 2 \cdot 1}}}$$ The $\color{#AA0000}{\text{colored}}$ terms can be cancelled to given what you have on your numerator.

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    Just FYI, you can use `\color{#rrggbb}{}` to customize your coloring in RGB mode so it's not so blinding and... more aesthetically pleasing. :-) – 2017-01-21