Combinations

Suppose four people $A,B,C,D$ must have three of them selected to serve on a committee. Unlike a permutation, here order doesn't matter — $A,B,C$ is exactly the same selection as $B,A,C$ or $C,A,B$. Listing every distinct group gives $ABC,\,ABD,\,ACD,\,BCD$: 4 ways of choosing 3 out of 4 people. Selecting 2 out of 4 similarly gives 6 distinct pairs. This count — the number of combinations of $n$ different objects taken $r$ at a time, order irrelevant — is denoted $^nC_r$; so $^4C_3=4$ and $^4C_2=6$.

Relating $^nC_r$ to $^nP_r$. Each combination of 3 objects can itself be arranged internally in $3!$ ways, so the permutations of 4 objects taken 3 at a time must be $^4C_3\times3!={}^4P_3$. In general,

Permutation vs. combination — side by side.