When selecting people to serve on committees, we need to be careful whether we use permutations or combinations (or both). Typically, if the committee members are ranked, such as chairman, vice chairman, recorder, we must use permutations. So if we want to know how many ways to select a ranked committee of 3 from 12 people, we would write

P(12,3) = 1320 ways

If the committee members are not ranked (they are simply members), then we use combinations. For example, the number of ways to choose a committee of three from 12 potential members is

C(12,3) = 220 ways

On some committees, there are ranked members and unranked members. For instance, if we had 12 people to choose from for a committee of 4 with a chairman and vice chairman, we would think of this using a slot diagram. One slot involves choosing the chairman and vice chairman and the other slot involves choosing the other two members. The number of ways to choose the chairman and vice chairman is

P(12,2) = 132

Since the other two members are not ranked, we choose them from the remaining 10 members,

C(10,2) = 45

Multiplying these numbers gives

P(12,2) ⋅ C(10,2) = 132 ⋅ 45 = 5940

In this case we need to use permutations and combinations since the committee contains ranked and unranked members.

Now let’s look at a tricky problem:

**Problem** Twelve people are to be distributed across three committees of four people each. How many ways is there to do this?

**Solution** Let’s think of this with a slot diagram where each slot corresponds to choosing a committee.

When choosing the first committee, we are choosing four people from twelve so the first slot becomes

Once we have chosen the first committee, there are eight people top choose from for the second committee:

Finally, we choose the third committee from the remaining people:

Putting in the correct values for each combination gives

This gives a total of 34,650 possible committees. However, we have overlooked something.

Using a slot diagram builds in an ordering issue. In counting the committees, we have counted different rearrangements as being different. In other words, this number counts C1, C2, C3 as being different from C3, C2, C1…and also different from C2, C1, C3. If we had assigned each committee a different task, we would want to rearrange the committees among the tasks and count those rearrangements.

In our problem, each committee is equal so we do not want the rearrangements to count. Since there are 3 ·2 · 1 = 6 ways to rearrange the committees, the total number of committees is