1. First we do the problem just as though it didn't matter whether John and
Marie were on the committee together or not.  

2. Then we'll calculate the number of ways John and Marie could be on the
committee together.

3.  Then we'll subtract the result of 2 from the result of 1.

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1. 

We choose the 3 girls:

That's "8 girls Choose 3" = 8 Choose 3 = 8C3 = 



For each of those 56 ways we can choose the 3 girls, we choose the 2 boys:

That's "6 boys Choose 2" = 6 Choose 2 = 6C2 = 

 

So the number of committees with no regard to whether or not John and Marie are
together is (56)(15) = 840 ways  

2.

We choose John and Marie to be together on the committee in 1 way.  

We choose the other 2 girls from the 7 girls besides Marie:

That's "7 boys Choose 2" = 7 Choose 2 = 6C2 = 

  

For each of those 21 ways to choose the other 2 girls, we can choose the 1 other
boy from the other 5 boys besides John.

That's 5 boys choose 1 = 5C1 =  = 5

So the number of committees with John and Marie together is (1)(21)(5) = 105
ways.

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3. Subtracting the result of 2 from the result of 1:

840 - 105 = 735 ways.

Edwin