# Why does $\binom{10}{7} = \frac{10!}{(10-7)!7!}$

We just learned that: $\dbinom{10}{7}= \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4}{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}$, so that:

If you throw a dice 10 times, the probability of getting $6$, $7$ of the times is: $\dbinom{10}{7} \times {\frac{1}{6}}^7 \times {\frac{5}{6}}^3$, because $\dbinom{10}{7}$ will give us the number of different ways you can get $7$ "correct" out of $10$.

I wonder why this works. Why does $\dbinom{10}{7}$ work as it does? (In my search I stumbled upon this way of writing it: $\dbinom{10}{7} = \frac{10!}{(10-7)!7!}$. It is a little bit different, but maybe it is more correct?

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2022-07-25 20:42:53
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