Question 1098055
Out of 6 physicists and 6 chemists, a committee consisting of 3
 physicists and 2 chemists is to be formed. In how many ways
can this be done if: two particular physicists cannot be on 
the committee together?
<pre>
So there are 2 "restricted" physicists and 4 non-"restricted" physicists.

So the committee consists of either:

Case 1: 1 restricted physicist, 2 non-resticted physicists, and 2 chemists.  

Choose the 1 restricted physicist in 2C1 ways.  (2 choose 1)
Choose the 2 non-restricted physicist in 4C2 ways. (4 choose 2)
Choose the 2 chemists in 6C2 ways. (6 choose 2)

That's (2C1)(4C2)(8C2).

          OR

Case 2: 3 non-restricted physicists, and 2 chemists.  

Choose the 3 non-restricted physicist in 4C3 ways. (4 choose 3)
Choose the 2 chemists in 6C2 ways. (6 choose 2)

That's (4C3)(6C2).

Total for both cases:  (2C1)(4C2)(8C2) + (4C3)(6C2)

Edwin</pre></font></b>