Joined: 23 Jun 2006
, given: 0
first choose which three (out of 4) pairs, have a representative in the committee (4 ways to do that... you can think of it as which pair doesn't have....)
now each pair can send, independently, 1 out of 2 possible representative.
so we have 2*2*2 = 8 possible committees once the three pairs are chosen.
since choosing the three pairs, and choosing the committee out of the given three pairs can be done independently... jut multiply it altogether to get 32
which is the right answer.
alternatively (i.e. what i would do on test day) just count.
for pairs 1,2,3: we have aaa,aab,aba,abb,baa,bab,bba,bbb
for pairs 1,2,4: aaa,aab,aba,abb,baa,bab,bba,bbb
now you can see that for each choices of pair you have 8 options. you have 4 choices of pairs and get 32 options altogether.