# How many different 20-person committees can be formed…

Q)How many different 20 persons committees can be formed each containing at least 2 professors and at least 3 Associate Professor from a set of 10 professors and 42 Associate Professors.

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Declarations of urgency are usually counterproductive... I suggest you choose a descriptive title for your problem and maybe you include some of the confused thoughts you have? – t.b. Sep 11 '12 at 12:22
try looking up nPr and nCr – yiyi Sep 11 '12 at 12:22
@MaoYiyi I believe searching for npr and ncr won't help much. Npr will bring up National Public Radio, for example. How about combinations and permutations? – Graphth Sep 11 '12 at 12:26
@Graphth sorry, I have just used the formula so much, that I forgot that someone may not understand what I wrote. Thanks, for keeping me on my toes. – yiyi Sep 11 '12 at 12:28
This is a very helpful site and I believe explains what you need better than wikipedia. mathsisfun.com/combinatorics/combinations-permutations.html – yiyi Sep 11 '12 at 12:29

Since you have to choose 20 persons out of a total of 52 persons and there are only 10 professors, so any combination of 20 people from the lot will contain at least 10 associate professors. Therefore what you have to do is find those combinations of 20 persons which have at least 2 professors. Total number of ways of selecting a group of 20 persons is $$52 \choose 20$$
Now we find the number of groups which have less than 2 professors, i.e., have either 0 or 1 professor in the group of 20 persons. Total number of such combinations are $${42 \choose 20}{10 \choose 0} + {42 \choose 19}{10 \choose 1}$$
Therefore, the required number of ways is $${52 \choose 20} - {42 \choose 20} - 10{42 \choose 19}$$