# Combination Problem

• August 14th 2008, 08:39 PM
RedBarchetta
Combination Problem
Four officers must be chosen from a high school committee of eight freshmen and five sophomores, with exactly two officers to be chosen from each class. in how many ways can these officers be chosen?

Okay, using this formula:

http://www.mathwarehouse.com/probabi...on-formula.gif

I came up with 715. With n=13; r=4. Although, this is clearly wrong.

How do I add in the element of specifically two members from each category, freshmen and sophomore, into this equation?

Thanks!
• August 14th 2008, 09:05 PM
mr fantastic
Quote:

Originally Posted by RedBarchetta
Four officers must be chosen from a high school committee of eight freshmen and five sophomores, with exactly two officers to be chosen from each class. in how many ways can these officers be chosen?

Okay, using this formula:

http://www.mathwarehouse.com/probabi...on-formula.gif

I came up with 715. With n=13; r=4. Although, this is clearly wrong.

How do I add in the element of specifically two members from each category, freshmen and sophomore, into this equation?

Thanks!

${8 \choose 2} \, {5 \choose 2} =$ ....
• August 14th 2008, 10:39 PM
RedBarchetta
....Thanks. My friend says the answer is 1,120 in his book yet this doesn't seem right to me.

$
\left( {\begin{array}{*{20}c}
8 \\
2 \\

\end{array} } \right)\left( {\begin{array}{*{20}c}
5 \\
2 \\

\end{array} } \right) = \frac{{8!}}
{{2!(8 - 2)!}} \cdot \frac{{5!}}
{{2!(5 - 2)!}} = \frac{{8!}}
{{2!6!}} \cdot \frac{{5!}}
{{2!3!}} = 28 \cdot 10 = 280
$