When doing combinations the number is it the number of outcomes the the number needed like 3 out of 5 would be nCr (5,3)?

$${\left({\frac{{\mathtt{5}}{!}}{{\mathtt{3}}{!}{\mathtt{\,\times\,}}({\mathtt{5}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)} = {\mathtt{10}}$$          YES  i Just tried it to check