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I have written in a textbook that $$\cfrac{n(n-1)(n-2)\times\cdots\times(n-r+1)}{r!}\tag{1}$$ $$=\cfrac{n(n-1)(n-2)\times\cdots \times 2 \times 1}{r!(n-r)(n-r-1)\cdots \times 2 \times 1}\tag{2}$$ $$=\cfrac{n!}{r!(n-r)!}\tag{3}$$

I understand completely how $(3)$ follows from $(1)$.

But I don't understand how $(2)$ follows from $(1)$. I tried multiplying top and bottom of $(1)$ by $(n-r)(n-r-1)$ which gives $$\cfrac{n(n-1)(n-2)\times\cdots\times(n-r+1)(n-r)(n-r-1)}{r!(n-r)(n-r-1)}$$ but this gets me nowhere.

Since it was written in textbook all in one step as shown above; my guess is that it must be something simple I am missing.

Any help will be greatly appreciated.

BLAZE
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    Well about (1) to (2) he just multiplied denominator and numerator by $(n-r)!$ – kingW3 Nov 06 '15 at 01:42
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    What you thought "got you nowhere" was actually getting you somewhere. Keep multiplying more terms (I mean (n - r - 2), (n - r - 3) and so on) and you'll get (2) eventually. – Tunococ Nov 06 '15 at 01:58
  • @Tunococ Would you please show me the steps to do this in an answer as I have tried and can't get it to work? – BLAZE Nov 06 '15 at 02:03

3 Answers3

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The recursive definition of $n!$ is that $n! = n(n-1)!$. We can extend this past $n-1$ to get:

$$n! = n(n-1)(n-2)!$$ $$n! = n(n-1)(n-2)(n-3)!$$ $$\dots$$ $$n! = n(n-1)(n-2) \dots (n-r+1)(n-r)!$$

Dividing the last equation by $(n-r)!$ gives $$n(n-1)(n-2)\dots(n-r+1) = \frac{n!}{(n-r)!}$$

Michael Biro
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As you said, you got $$ \frac{n(n - 1)(n - 2) \cdots (n - r - 1)}{r!(n - r)(n - r - 1)} $$ by multiplying $n - r$ and $n - r - 1$ to the numerator and the denumerator in (1). Let's see what happens when you keep multiplying the "next term" until you reach $1$.

Multiply by $n - r - 2$: $$ \frac{n(n - 1)(n - 2) \cdots (n - r - 1)(n - r - 2)}{r!(n - r)(n - r - 1)(n - r - 2)} $$ Then multiply by $n - r - 3$: $$ \frac{n(n - 1)(n - 2) \cdots (n - r - 1)(n - r - 2)(n - r - 3)}{r!(n - r)(n - r - 1)(n - r - 2)(n - r - 3)} $$ This process can continue until you reach $n - r - k = 1$ for some $k$. In other words, I can write the expression, with the help of $\cdots$, as $$ \frac{n(n - 1)(n - 2) \cdots 1}{r!(n - r)(n - r - 1)\cdots 1}. $$ Just be aware that when you use $\cdots$, you may choose to stop whenever you think it is clear enough for your reader, so the expression is probably not unique. For example, you could write this last expression as $$ \frac{n(n - 1)\cdots 1}{r!(n - r)(n - r - 1)\cdots 1} \quad \text{ or } \quad \frac{n(n - 1)(n - 2)\cdots 2 \cdot 1}{r!(n - r)(n - r - 1)\cdots 2 \cdot 1} $$ and I would still understand you.

Tunococ
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Using the product notation might help.

$\begin{array}\\ \frac{n(n-1)(n-2)\times\cdots\times(n-r+1)}{r!} &=\frac{\prod_{k=n-r+1}^n k}{r!}\\ &=\frac{\prod_{k=n-r+1}^n k}{r!}\frac{\prod_{k=1}^{n-r} k }{\prod_{k=1}^{n-r} k }\\ &=\frac{\prod_{k=1}^n k}{r!}\frac{1 }{(n-r)! }\\ &=\frac{n!}{r!(n-r)!}\\ \end{array} $

marty cohen
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