Permutations/Combinations question

**Runty** · March 4th 2010, 08:14 AM

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?

Please show your work.

**Plato** · March 4th 2010, 08:29 AM

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?
Please show your work.

By all means, do show your work.

**Runty** · March 7th 2010, 10:19 AM

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

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

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

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

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

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

**Plato** · March 7th 2010, 10:32 AM

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

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

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

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

You are about 1,2,& 4.
For #3 I have $\binom{10}{4}\binom{4}{4}$ $+\binom{10}{5}\binom{4}{3}$ $+\binom{10}{6}\binom{4}{2}$

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