Is that meant to be something like "If you want a perm at the hairdressers, you have to order, fi you want a comb, you don't"? It took me about a minute to make sense of it.
Personally I think remembering "permutation=ordered, combination=unordered"
Also, a combination is the permutation on r!, or (n!/(n-r)!)/r!, which is n!/[r!(n-r)!].
Um. Here is a lame example: I recently made a poll in the School forum which had options in groups of two universities. Eg: a is a univeristy, B is another: one option was AB. Now, there are 23 universities in Australia, I think, and we are taking them TWO at a time. The amount of possible permutations is then 23^P_2, or 23!/(23-2)!=22*23=506. HOWEVER, order doesn't matter. Here, the combination (uni A, uni B) is exactly the same as the combination (uni B, uni A). So we must divide our number of permutations by r!, which is 2!, or 2. Thus the answer is 253.
We could skip the permutation step and do this: 23^C_2,
23!/(2*21!)=23*22/2=11*23=253.
Sorry. I did that ebcause I think comittee and people questions are boring.