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Regarding my previous post , I'll repeat the question

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 if the committee must contain at least three Math teachers.

I know it could be solved like this

$\binom{4}{3}\binom{5}{2} + \binom{4}{4} \binom{5}{1} = 45 $ Ans

I wanted to know How I would solve this the other way round. Like for example if it was for at least 1 math teacher it would be

$\binom{9}{5} - \binom{5}{5} $ how would I use the same method but instead calculating for at least 1 I would be calculating for at least 3 ?

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    @joriki: Actually, if I extend the problem to include another department (Chemistry) with$3$teachers, then "At least one math teacher" becomes $\mbox{All possibilities} - \mbox{No Math teachers}$ which is $\binom{12}{5} - \binom{4}{0}\binom{8}{5}.$ As you can see, the focus here is with the $\binom{4}{0}$ because that's what says "No math teachers."2012-08-07

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If I understand you correctly, you want to use that sort of subtraction technique, where you overcount and then take out the complement of the desired probability, to solve the "at least 3" math teacher case.

Then you might do ${9 \choose 5} - \left( {\color{#10C}{5 \choose 5}} + \color{#070}{{5 \choose 4}{4 \choose 1}} + \color{#C01}{{5 \choose 3}{4 \choose 2}}\right)$, where $\color{#10C}{blue}$ counts the number of ways of choosing 5 English people, $\color{#070}{green}$ counts the number of ways of choosing 4 English and 1 mathie, and $\color{#C01}{red}$ counts the number of ways of choosing 3 English and only 2 mathies.

Of course, this is nothing more than saying that ${9 \choose 5} = {5 \choose 5} + {5 \choose 4}{4 \choose 1} + { 5 \choose 3}{4 \choose 2} + {5 \choose 2}{4 \choose 3} + { 5 \choose 1}{4 \choose 4}$.