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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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It's hard for me to see exactly what part of this you're finding difficult. You know how to calculate the number of combinations with $0$, $3$ and $4$ math teachers, so it seems you would also be able to calculate the number of combinations with $1$ and $2$ math teachers. You also know how to calculate the overall number of combinations, and that you can substract the numbers for some cases from that to get the sum of the numbers for the remaining cases. If you know all that, what's keeping you from subtracting the numbers for $0$, $1$ and $2$ math teachers from the total number? – joriki Aug 7 '12 at 0:44
@joriki. I tried using that approach $\binom{9}{5} - ( \binom{5}{5} + \binom{6}{5} + \binom{7}{5} ) $ but I dont get the correct answer – MistyD Aug 7 '12 at 0:49
This is why it's a good idea to show what you've tried; then people can specifically point out what went wrong instead of guessing what the answer should focus on. – joriki Aug 7 '12 at 0:59
@MistyD: It is good to be aware of both the direct and "inverse" way of counting. In this case, the inverse way requires more calculation, so one would not use it. But fairly often it saves a lot of effort. – André Nicolas Aug 7 '12 at 1:06
@MistyD: I think you meant $$\binom{9}{5} - \binom{4}{0}$$ for the number of ways to select at least one math teacher. What you've written says the number of ways of having at most four English teachers. – ladaghini Aug 7 '12 at 1:17
up vote 1 down vote accepted

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 ${9 \choose 5} = {5 \choose 5} + {5 \choose 4}{4 \choose 1} + { 5 \choose 3}{4 \choose 2} + {5 \choose 2}{4 \choose 3} + { 5 \choose 1}{4 \choose 4}$.

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