First we'll calculate as though the committees are ordered: committee A,
committee B, and committee C. Then we'll "unorder" the committees by dividing
by 3!, the number of orderings of the 3 committees.
We can choose the 11 people for committee A in 33C11 = 193536720 ways.
We can then choose the 11 people for committee B in 22C11 = 705432 ways.
Then the 11 remaining people make up committee C in 11C11 = 1 way.
That's (33C11)(22C11)(11C11) = 136526995463040
Now we must divide that by 3!=6 because the committees can be permuted in 3!
= 6 ways,
That's (33C11)(22C11)(11C11)/3! = 22754499243840
Edwin
