1. ## Permutations/Combinations question

Given a group of 4 computer science professors and 10 of their TAs:

(a) how many different 6 person soccer teams can be made?

(b) how many different 5 person teams can be made that contain at least 1 professor?

(c) how many different 8 person teams can be made that contain at least 2 professors and at least 4 TAs?

(d) how many different 4 person teams can be made that contain exactly 2 TAs?

2. Originally Posted by Runty
Given a group of 4 computer science professors and 10 of their TAs:
(a) how many different 6 person soccer teams can be made?
(b) how many different 5 person teams can be made that contain at least 1 professor?
(c) how many different 8 person teams can be made that contain at least 2 professors and at least 4 TAs?
(d) how many different 4 person teams can be made that contain exactly 2 TAs?
By all means, do show your work.

3. Okay, here are the answers I've gotten so far, but I want to make sure if I'm doing them right.

a) $\displaystyle C(14,6)=\frac{14!}{6!8!}=3003$

b) $\displaystyle C(14,5)-C(10,5)=\frac{14!}{5!9!}+\frac{10!}{5!5!}=2002-252=1750$

c) $\displaystyle C(14,8)-4[C(10,7)]-C(10,8)=\frac{14!}{8!6!}-4\frac{10!}{7!3!}-\frac{10!}{8!2!}=3003-4(120)-45=2478$

d) $\displaystyle \frac{10\times 9}{2}\times \frac{4\times 3}{2}=45\times 6=270$

The ones I'm not sure about are the third and fourth ones. Can you confirm if these are correct?

4. Originally Posted by Runty
Okay, here are the answers I've gotten so far, but I want to make sure if I'm doing them right.

a) $\displaystyle C(14,6)=\frac{14!}{6!8!}=3003$

b) $\displaystyle C(14,5)-C(10,5)=\frac{14!}{5!9!}+\frac{10!}{5!5!}=2002-252=1750$

c) $\displaystyle C(14,8)-4[C(10,7)]-C(10,8)=\frac{14!}{8!6!}-4\frac{10!}{7!3!}-\frac{10!}{8!2!}=3003-4(120)-45=2478$

d) $\displaystyle \frac{10\times 9}{2}\times \frac{4\times 3}{2}=45\times 6=270$
For #3 I have $\displaystyle \binom{10}{4}\binom{4}{4}$$\displaystyle +\binom{10}{5}\binom{4}{3}$$\displaystyle +\binom{10}{6}\binom{4}{2}$