Question 870213
If one particular physicist must be on the committee, then you really only have 3-1=2 slots left for physicists. So you'll have 7-1 = 6 physicists left to pick from.



There are 5 nCr 2 = 10 ways to pick 2 mathematicians (from a pool of 5)



There are 6 nCr 2 = 15 ways to pick 2 physicists (from a pool of 6)



Side Note: I'm using the <a href="http://www.mathwords.com/c/combination_formula.htm">combination formula</a> to compute nCr



So there are 10*15 = 150 ways to form a 4 person committee (2 mathematicians and 2 physicists)



By extension, there are also 150 ways to form a 5 person committee (2 mathematicians and 3 physicists) with one physicist slot taken. This is because that particular physicist who must be on the committee doesn't change the count (you'll have 1*150 = 150)