Proving identity with factorials: $K\cdot(K - 1) \cdot(K - 2) \cdot \dots\cdot (K - N + 1) = K!/(K-N)!$

Question

I was looking at some formulas in one of my courses and I see the following equality:

$$K\cdot(K - 1)\cdot(K - 2)\cdots(K - N + 1) = \frac{K!}{(K-N)!}$$

Can't see a way to figure out how these are equal. Thank you.

Answer 1

Note that $$(k-n)!=(k-n)(k-n-1)(k-n-2) \dots 1$$, and that $$k!=k(k-1)(k-2) \dots (k-n)(k-n-1) \dots 1$$.

Since you are dividing $k!$ by $(k-n)!$, erase the terms they have in common. You get $k(k-1)(k-2)\dots(k-n+1)$.

Answer 2

$K!=K*(K-1)*(K-2)*....*(K-N+1)*(K-N)*(K-N-1)*.....*3*2*1$...

$(K-N)!=(K-N)*(K-N-1)*.....*3*2*1$...

Thus...

Answer 3

$K!=K\cdot(K-1)\cdot(K-2)\cdot\cdot\cdot (K-N+1)\color{blue}{(K-N)(K-N-1)\cdot\cdot\cdot3\cdot2\cdot1}$

And

$\color{blue}{(K-N)!=(K-N)\cdot(K-N-1)\cdot(K-N-2)\cdot\cdot\cdot3\cdot2\cdot1}$

So $ K\cdot(K-1)\cdot(K-2)\cdot\cdot\cdot (K-N+1)= \frac{\color{red}{K!}=K\cdot(K-1)\cdot(K-2)\cdot\cdot\cdot (K-N+1) \color{blue}{\require\bcancel{(K-N)(K-N-1)\cdot\cdot\cdot3\cdot2\cdot1}}}{\color{red}{(K-N)!} \color{blue}{(K-N)\cdot(K-N-1)\cdot(K-N-2)\cdot\cdot\cdot3\cdot2\cdot1}}$

Since you are dividing by $(k-n)!$, you should cancel the $\color{blue}{\text{common terms }}$.

Hope it helps