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I'm taking an online probability and statistics course through edX called Fat Chance: Probability from the Ground Up, but I've never taken anything other than high school math (the usual fodder — algebra, geometry) and a couple of college calculus courses. If any of those covered anything remotely related to combinatorial mathematics, I must have missed it. Needless to say, I need an answer that breaks this equation down step by step because clearly some basic building blocks are missing. If you can deliver an answer in simple, yet thorough language that I can understand, many thanks to you in advance.

BTW, the course I mentioned does allow you to post questions, but not having done so before, I don't know how responsive the staff is to such questions currently (since I think the course has been out for a couple of years). So, I thought I'd post this question here as a way to test out this Stack Exchange, which I just joined today. If I get an answer from edX that bests anything posted here, I'll add it.

Lisa Beck
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If $n,k \in \mathbb{N}$ such that $n \geq k$, one defines

$$\binom{n}{k} := \frac{n!}{k!\cdot (n-k)!},$$

("$n$ choose $k$"), whereby

$$n ! := n\cdot (n - 1)!, \ \ \ 0!:= 1.$$

Note that $n!$ is simply the product of all natural numbers up to $n$, so for $n \geq 1$ we have

$$n! = \prod_{i = 1}^n i = 1 \cdot 2 \cdot \ldots \cdot n.$$

From the definitions we now conclude

\begin{align} \binom{9}{4} \cdot \binom{5}{3} &= \frac{9!}{4!\cdot 5!} \cdot \frac{5!}{3!\cdot 2!} \\ &= \frac{(9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1) \cdot (5 \cdot 4 \cdot 3 \cdot 2 \cdot 1)}{(4 \cdot 3 \cdot 2 \cdot 1) \cdot (5 \cdot 4 \cdot 3 \cdot 2 \cdot 1) \cdot (3 \cdot 2 \cdot 1) \cdot (2 \cdot 1)} \\ &=126 \cdot 10 \\ &= 1260.\end{align}

Jan
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  • What a fantastic answer. Extra points for neatness and style, but I don't understand why the denominator becomes 4!⋅5! for the first ... you know what? Nevermind. It is clearly spelled out for me with this: n! / k!⋅(n−k)! Learning has occurred! Thank you for so clearly explaining this to me. – Lisa Beck Jul 31 '19 at 09:00
  • @LisaBeck For the denominator just use the definition. In $\binom{9}{4}$ we have $n = 9$ and $k = 4$. So as written above, since $n - k = 5$, it follows $\binom{n}{k} = \binom{9}{4} = \frac{n!}{k!\cdot (n - k)!} = \frac{9!}{4! \cdot 5!}$. – Jan Jul 31 '19 at 09:03
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    I just did the calculations to make sure I really understood this. It worked perfectly! Math is a joy when you finally understand it. Thanks again for your helpful, speedy response to my question! – Lisa Beck Jul 31 '19 at 09:12