# Double counting in combinatorial formulae.

I've just started learning combinatorics, and have the following problem in reasoning.

One of the formulae for ${n \choose k}$ is: $${n \choose k} =\frac{n!}{(n-k)!k!}= \frac{n \ \times \ (n-1) \ \times \ (n-2) \ \times \ ... \ \times \ (n-k+1)}{k \ \times \ (k-1) \ \times \ (k-2) \ \times \ ...\ \times \ 1}$$

My understanding is that we need the denominator to avoid the problem of counting multiple times. For example, this would help avoiding counting the sets {1,2,3,4} and {2,1,4,3} twice, if we want to count the number of 4-elements subsets of n elements.

My problem lies in analyzing other combinatorial situations, where I often don't know if I need to do something similar to the above. Am I correct to assume that every time we actually want all $k!$ permutations to be counted (i.e. we don't mind and actually need both {1,2,3,4} and {2,1,4,3} in the example above to be counted), then we don't have to worry about double counting?

You are right. When you want to arrange $k$ from total $n$ elements and order is counted, there are $$A_n^k=\frac{n!}{(n-k)!}={n\choose k}\cdot k!$$
ways to do this! You can take it as first choosing $k$ elements without considering the order, then you permute them in all possible ways.