Question 1176410
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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