there are 20 teachers, of these are 8 maths teachers, 6 history teachers, 4 physics teachers and 2 geography teachers.

how many ways can the teachers be chosen if there are to be at least 2 maths teachers?

i know the correct way to do this is: total number of combinations - combinations if there are no maths teachers - combinations if there is 1 math teacher

or even: combination if there are 2 maths teachers + combinations if there are 3 maths teachers + combinations if there are 4 maths teachers

but, what's wrong with the following approach? it seems to make sense: first, choose 2 from the 8 maths teachers, and then choose the other 2 from the remaining 18 teachers

$\displaystyle ^8C_2\cdot^{18}C_2$

this gives a completely dfiferent answer

(Edited something i completely mistakenly typed in)