Mrs. Welch and her 5 children are shopping at a local grocery store. Each of the children will be allowed to select one box of cereal for their own from the 11 different boxes of cereal available (there are no two the same). In how many ways can the selections be made?

Explanation:

Let's call the children A, B, C, D, E for the sake of simplicity. Also, let's have the children select in alphabetical order.

We started at 11 and counted down by 1 until we had all the children selecting.

Multiply out these values: 11*10*9*8*7 = 55440

There are 55440 different permutations possible. If order mattered, then we would stop here. However, order does not matter because the cereals aren't ranked and their position doesn't matter. All that matters is what fills the grocery cart. In other words, imagine the kids jumbling up the boxes as they push around the cart, so the order would be irrelevant.

Consider a group of 5 items. There are 5! = 5*4*3*2*1 = 120 different ways to arrange the five items in this group. So for any 5 cereals picked, there are 120 ways to order them. But again, order doesn't matter, so we have to correct for this over-counting. We do this by dividing by 120

55440/120 = 462

If you want to use a formula, then plug n = 11 and r = 5 into the combination formula below and simplify.

n C r = (n!)/(r!(n-r)!)

11 C 5 = (11!)/(5!*(11-5)!)

11 C 5 = (11!)/(5!*6!)

11 C 5 = (11*10*9*8*7*6!)/(5!*6!)

11 C 5 = (11*10*9*8*7)/(5!)

11 C 5 = (11*10*9*8*7)/(5*4*3*2*1)

11 C 5 = 55440/120

11 C 5 = 462

We get the same answer.