# Set operations

I have two committees A and B, with |A| = 3 = |B|, out of a group of 8

people such that at most one person can serve on both committees. How many ways we can form the committees A and B? Please show me how it's done.

https://brainmass.com/math/group-theory/set-operations-committee-form-two-groups-people-205373

## SOLUTION This solution is **FREE** courtesy of BrainMass!

In this question, we need to consider two cases. Notice that the question says that "at most" one person can serve on both committees. This means that there could be zero or one person on both committee. We need to separate the two cases and consider them differently.

(1) For the case that no person serves on both committee

Then the selection process is as follows: we would first select three people out of the eight for committee A. We would then select another three people out of the remaining five people (eight people minus three) for committee B.

The number of ways to select committee A is 8C3 = 8! / [3! * (8 - 3)!] = 56

where 8! = 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 and 8C3 is read "8 chooses 3". The formula for nCr = n! / [r! * (n - r)!]

Similarly, the number of ways to select committee B is 5C3 = 10

Under this case, the total number of selecting both committee A and B is 56 * 10 = 560

(2) For the case that there is one person serves on both committees. We would do the following process. Pick one person out of the eight to be the person who serves on both committees. We would then select two people out of the seven for committee A. Lastly, we would select another two people out of the remaining five people for committee B.

The number of ways to select one person out of eight = 8

The number of ways to select two people for committee A is 7C2 = 21

The number of ways to select two people for committee B is 5C2 = 10

Under this case, the total number of ways to select committee A and B is 8 * 21 * 10 = 1680

Then combine (1) and (2), the total number of selecting committee A and B is 560 + 1680 = 2240

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