# Combination/Permutation

• Sep 11th 2007, 12:50 PM
aikenfan
Combination/Permutation
A panel of judges is to consist of fourteen women and five men. A list of potential judges includes sixteen women and seven men. How many different panels could be created from this list?
(A) 360
(b) 5040
(c) 2520
(d) 158

I am pretty sure that you use combination, but i can seem to get any of these numbers....could anyone please help me with this problem?
• Sep 11th 2007, 01:04 PM
Jhevon
Quote:

Originally Posted by aikenfan
A panel of judges is to consist of fourteen women and five men. A list of potential judges includes sixteen women and seven men. How many different panels could be created from this list?
(A) 360
(b) 5040
(c) 2520
(d) 158

I am pretty sure that you use combination, but i can seem to get any of these numbers....could anyone please help me with this problem?

Yes! you indeed have to use combinations. out of the 16 women, you must choose 14. out of the 7 men, you must choose 5. so the answer is "16 choose 14 * 7 choose 5"...or in math language: $\displaystyle {16 \choose 14}\cdot{7 \choose 5}$

• Sep 11th 2007, 01:06 PM
aikenfan
So would it be 16 C 14?
and 7 C 5?
• Sep 11th 2007, 01:16 PM
Jhevon
Quote:

Originally Posted by aikenfan
So would it be 16 C 14?
and 7 C 5?

Yes. $\displaystyle _nC_r = C(n,r) = {n \choose r}$ they are all different notations for the same thing
• Sep 11th 2007, 01:25 PM
aikenfan
I get 120 and 21?
• Sep 11th 2007, 02:01 PM
Jhevon
Quote:

Originally Posted by aikenfan
I get 120 and 21?

you multiply them. that's what * means
• Sep 11th 2007, 03:42 PM
aikenfan
Thank you very much