SOLUTION: Question: There are 5 physicists, 4 chemists, and 3 mathematicians. In how many ways can the committee of 6 be chosen so that AT LEAST 3 members of the committee are physicists?

Question 1176410: Question: There are 5 physicists, 4 chemists, and 3 mathematicians. In how many ways can the committee of 6 be chosen so that AT LEAST 3 members of the committee are physicists?
My answer: C(5,3) = 10.
Once the three physicists are selected, since it is AT LEAST, the remaining 2 physicists can still be selected which makes available candidates for the remaining 3 positions n = 9
We need another 3. so C(9,3) = 84.
Therefore, the number of ways the committee could be chosen is 10 X 84 = 840
But the book says 462.
What am I doing wrong?
Thanks for your help.
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
12C6 are the number of possible ways to choose 6
at least 3 physicists would be
3: 5C3*7C3 for the other 3 which is 10*35 or 350. Consider once you have 5C3 in the numerator, you need another 7 on top and 3 on the bottom.
4, 5C4*7C2 which would be 105. Notice how the two change
5 5C5*7C1, or 7
add to 462.
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The candidates for the other 3 positions can't be physicists so the choices are 7C3.
And at least 3 members means 4 and 5 can be physicists as well.
You are assuming the first 3 are physicists, and yes, that leaves 9. But if only 3 are required, then it leaves 7, not 9 people, since a 4th physicist won't be allowed.
If 4 are required, it leaves 7 as well.
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