# What is the exact connection between Binomial coefficients and Factorials

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?

• Welcome to Mathematics Stack Exchange. Binomial coefficients are ratios of factorials – J. W. Tanner Mar 2 at 19:54

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).
$$\dbinom{n}{k}$$ 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.