# Math Help - combinations

1. ## combinations

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
$^8C_2\cdot^{18}C_2$
this gives a completely dfiferent answer

(Edited something i completely mistakenly typed in)

2. ## Re: combinations

Counting combinations means that the order in which the elements are selected doesn't matter. However, by choosing two math teachers first, then 2 teachers (from a pool that include other math teachers), you are adding an order. First you chose two, then you possibly choose more. By adding an ordering to the choices, you are no longer counting combinations.

3. ## Re: combinations

Originally Posted by SlipEternal
Counting combinations means that the order in which the elements are selected doesn't matter. However, by choosing two math teachers first, then 2 teachers (from a pool that include other math teachers), you are adding an order. First you chose two, then you possibly choose more. By adding an ordering to the choices, you are no longer counting combinations.
Thanks! i wasn't able to see that until i wrote down some of the possible outcomes of using this approach with a smaller sample space.

to be clear (this is mostly just for my own benerit lol), what happens is that in the first part i may have chosen math teacher A and math teacher B, then in the second part i chose math teacher C and geography teacher A. In another draw, i choose maths teacher B and C in the first part, then choose maths teacher A and geography teacher A. These 2 draws are the same but are counted as different, because of the arrangemnt as slipeternal said.

4. ## Re: combinations

Originally Posted by muddywaters
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?
You must tell how many are to be chosen.

5. ## Re: combinations

Originally Posted by Plato
You must tell how many are to be chosen.
Oh yes sorry it was 4, missed that out.

6. ## Re: combinations

Originally Posted by muddywaters
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 (four) teachers be chosen if there are to be at least 2 maths teachers?
$\binom{20}{4}-\left(\binom{12}{4}+8\binom{11}{3}\right)$ WHY?

7. ## Re: combinations

Originally Posted by Plato
$\binom{20}{4}-\left(\binom{12}{4}+8\binom{11}{3}\right)$ WHY?
Rather, $\binom{20}{4}-\left(\binom{12}{4}+8\binom{12}{3}\right)$.
Because,
Originally Posted by muddywaters
total number of combinations - combinations if there are no maths teachers - combinations if there is 1 math teacher