User:IssaRice/Summary of counting techniques

Let $A$ be a set with $n$ elements, and let $B$ be a set with $m$ elements.

Description	Set representing counting problem	number of ways to count
Pick $k$ things from $A$ with replacement	${(a_{1}, \dots, a_{k}) : a_{1}, \dots, a_{k} \in A}$	$n^{k}$
	${{a_{1}, \dots, a_{k}} : a_{1}, \dots, a_{k} \in A}$	$\sum_{i = 1}^{k} (\binom{n}{i})$
	${f : A \to N ∣ \sum_{a \in A} f (a) = k}$
	${{a_{1}, \dots, a_{n}} : a_{1}, \dots, a_{n} \in A}$	$\sum_{i = 1}^{n} (\binom{n}{i}) = 2^{n} - 1$ (a quick way to see this identity is that we want the power set without the empty set)
	${(a_{1}, \dots, a_{k}) : a_{1}, \dots, a_{k} \in A and all a_{i} distinct}$	$P (n, k) = \frac{n!}{(n - k)!} = n (n - 1) \dots (n - (k + 1))$
	${(a_{1}, \dots, a_{n}) : a_{1}, \dots, a_{n} \in A and all a_{i} distinct}$	$P (n, n) = n!$
	${{a_{1}, \dots, a_{k}} : a_{1}, \dots, a_{k} \in A and all a_{i} distinct}$	$(\binom{n}{k}) = P (n, k) / (k!) = \frac{n!}{k! (n - k)!}$
	${(a, b) : a \in A and b \in B}$	$n m$
	${{a, b} : a \in A and b \in B}$