1. First we do the problem allowing them to serve on the same committee.
2. Then we do a second problem requiring them to serve on the same committee.
3. Then we subtract the answer to (2) from the answer to (1).

1. That's 6 people choose 3 or 6C3 = 20
2. If those two are on the same committee, we choose the a third man
to serve with them.  There are 4 other men, so we can be on the same
committee 4 ways. That's 4 people choose 1 or 4C1 or 4 ways.

3. Answer:  20-4 = 16 ways. 

Checking:

Suppose the 6 people are Alan, Betty, Cathy, Donald, Edwin, and Flo.
Suppose Alan and Betty will not serve on the same committee. 

Here are all 16 committees. Notice that Alan and Betty are not 
together on any of them:

 1.  {Alan, Cathy, Donald}
 2.  {Alan, Cathy, Edwin}
 3.  {Alan, Cathy, Flo}
 4.  {Alan, Donald, Edwin}
 5.  {Alan, Donald, Flo}
 6.  {Alan, Edwin, Flo}
 7.  {Betty, Cathy, Donald}
 8.  {Betty, Cathy, Edwin}
 9.  {Betty, Cathy, Flo}
10.  {Betty, Donald, Edwin}
11.  {Betty, Donald, Flo}
12.  {Betty, Edwin, Flo}
13.  {Cathy, Donald, Edwin}
14.  {Cathy, Donald, Flo}
15.  {Cathy, Edwin, Flo}
16.  {Donald, Edwin, Flo}

Edwin