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 ?