There are 8 members of a club. Four members are to be chosen
so that it contains A or B but not both. In how many ways can
it be done?
I suppose the members are {A,B,C,D,E,F,G,H}
Here are two ways to work it.
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Method 1:
1. First we enumerate all possible committees without
restrictions as to whether A and B are together or not.
then
2. We enumerate all committees that contain both A and B.
then
3. We enumerate all committess that do not contain either A or B.
4. We subtract the results of 2 and 3 from the result of 1
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1. 8 choose 4 = 8C4 = 70
2. We choose 2 from {C,D,E,F,G,H} to go with A and B
6 choose 2 = 6C2 = 15
3. We choose 4 from {C,D,E,F,G,H}
6 choose 4 = 6C4 = 15
4. Answer = 70-15-15 = 40
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Another way to do it:
Method 2.
1. Choose the other three committee members besides A and B
2. Multiply by 2, the number of ways we can choose either A or B
to be on the committee with those 3.
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1. 6 choose 3 = 6C3 = 20
2. 2*20 = 40.
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The second way is shorter.
Edwin
