# Dealing with “at least” in Permutation

For the following question (which I pulled of the internet)

A five member committee is to be selected from among four Math teachers and five English teachers. In how many different ways can the committee be formed under the following circumstance?

A) Anyone is eligible to serve on the committee.

B) The committee must consist of $3$ Math teachers and $2$ English teachers.

C) The committee must contain at least three Math teachers.

D) The committee must contain at least three English teachers.

(Answer $126$, $40$, $45$, $81$)

How would I go about solving when the requirement is at least $3$ Math Teacher Any suggestions ? . I know that when it was $3$ Math and $2$ English teachers I simply took r=3 for math and r=2 for English in the Combination Formula.

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Since you know how to do it for exactly $3$ math teachers, you also know how to do it for exactly $4$ math teachers. There are only $4$ math teachers, so those are the only two possibilities; you just have to add them up.
I still dont get it when there are $3$ Math teachers the combination is 4 and when there are 4 math teachers the combination is 5 so when there are 2 and 1 English teacher(s) the combinations are 10 and 5 respectively – MistyD Aug 6 '12 at 7:01
Okay so for 3 Math and 2 English it will be $4 \times 10$ and when there are 4 Math and 1 English it will be $5 \times 5$. How do I figure out for at least 3 Math. I am still a bit confused – MistyD Aug 6 '12 at 7:14
@MistyD: I don't know how you got $5\cdot5$; there are only $4$ math teachers in all, so there's only one way to choose $4$ of them, not $5$ ways. On your second question: What other possibilities except for $3$ and $4$ math teachers do you see for the combination to include at least $3$ math teachers if there are $4$ math teachers in all? – joriki Aug 6 '12 at 7:17
@joriki sorry that was a misprint when there are 4 Math and 1 English Teacher then its $1 \times 5$ – MistyD Aug 6 '12 at 7:22