The two standard math notations do not work terribly well in text, but I think I have seen the "Choose" notation written as a combinatorial function like so: Comb(n,k) = n!/(k!*(n-k)!). The formula should look somewhat familiar even if the notation does not.

In this problem you want to choose two teachers from a group of five, written as: Comb(5,2) = 5!/(3!*2!) = (5*4*3*2*1)/{(3*2*1)*(2*1)} = 10.

Also, you want to choose four students from a group of ten, written as Comb(10,4) = 10!/(6!*4!) = (10*9*8*7*6!)/{6!*(4*3*2*1)} = 10*3*7 = 210.

(To get to that second to the last step I canceled out the 6! from the numerator and denominator. I canceled the 4*2 in the denominator with the 8 in the numerator. And finally, canceled the 3 in the denominator with the 9 in the numerator leaving a 3.)

Now, since you have 10 ways to get the teachers in a committee and 210 ways to get the students into a committee, you multiply to get the total ways a committee can be constructed, i.e. 2100 ways.

Two committees will be formed. That leaves 1 teacher and 2 students without committee membership. This is the case if it is presumed that no teacher and students can take more than 1 committee,