# Combinations

• Jul 21st 2009, 02:59 AM
noobonastick
Combinations

A committee of 7 politicians is chosen from 10 liberal members, 8 labor members and 5 independents. In how many ways can this be done so as to include exactly 1 independent and at least 3 liberal members and at least 1 labor member?
• Jul 21st 2009, 06:18 AM
Plato
What do you think the answer is?
What have you done? Where are you having trouble?
• Jul 21st 2009, 01:28 PM
noobonastick
I have honestly no idea where to start. Teacher didn't explain how to do these questions.
• Jul 21st 2009, 01:39 PM
Plato
Quote:

Originally Posted by noobonastick
A committee of 7 politicians is chosen from 10 liberal members, 8 labor members and 5 independents. In how many ways can this be done so as to include exactly 1 independent and at least 3 liberal members and at least 1 labor member?

$\sum\limits_{k = 0}^2 {{\binom{5}{1}}{\binom{8}{k+1}}{\binom{10}{5-k}}}$
• Jul 21st 2009, 01:54 PM
noobonastick
Sorry, are you sure you havent done anything wrong? My book says the answer is 73 080
• Jul 21st 2009, 02:05 PM
Plato
Quote:

Originally Posted by noobonastick
Sorry, are you sure you havent done anything wrong? My book says the answer is 73 080

Maybe you need to improve your calculation skills.
• Jul 21st 2009, 02:14 PM
noobonastick
This is what i calculated.

(5C1).(8C1 + 8C2 + 8C3).(10C5 + 10C4 + 10C3)
• Jul 21st 2009, 02:19 PM
Plato
Quote:

Originally Posted by noobonastick
This is what i calculated.

(5C1).(8C1 + 8C2 + 8C3).(10C5 + 10C4 + 10C3)

WHY? The answer I gave you is a sum not a product.
$\binom{5}{1}\binom{8}{1}\binom{10}{5}+\binom{5}{1} \binom{8}{2}\binom{10}{4}+\binom{5}{1}\binom{8}{3} \binom{10}{3}$