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