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The factorial n! is the number of ways, the set {1, . . . , n} may be ordered.

The binomial coefficient defines how many different ways there are to choose out of a group of n exactly k of them.

What is the connection between factorials and binomial coefficients?

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  • $\begingroup$ Welcome to Mathematics Stack Exchange. Binomial coefficients are ratios of factorials $\endgroup$ – J. W. Tanner Mar 2 at 19:54
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Here is the combinatorical approach:

Suppose that we choose $k$ out of $n$ elements where order does matter. We would have $n$ posibilities for the first choice, $n-1$ posibilities for the next and so on. So we have $$ n\cdot (n-1) \cdot (n-2) ... (n-k+1) = \frac{n!}{(n-k)!}$$ posibilities, but this is only when order does matter. When order doesn't matter, we would have to divide by the number of ways we can arrange the $k$ objects, which of course is $k!$. We thereby get that there is $$\frac{n!}{k!(n-k)!}$$ ways to choose $k$ elements from a set of $n$ elements (when order does not matter).

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The connection comes from the other definition of a binomial coefficient:

$\dbinom{n}{k}$ is the coefficient of $x^k$ in the expansion (by distributivity) of $$(1+x)^n=\underbrace{(1+x)(1+x)\dotsm(1+x)}_{n\text{ factors}}$$

Indeed, to obtain $x^k$, in each of all factors $1+x$, we have to choose $k$ times $x$ and $n-k$ times $1$, in all possibles ways.

Using this approach, we can prove, differentiating $(1+x)^n$ and comparing the expansion of the derivative $n(1+x)^{n-1}$ and the derivative of the expansion, we obtain the following recursion formula: $$\binom nk=\frac nk\binom{n-1}{k-1}$$ from which the usual formula for the value of $\binom nk$ in terms of factorials, is deduced.

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