# Thread: number of possibilities

1. ## number of possibilities

how many 3 person committees can be formed in a club with 8 members?

2. Hello,

This is a problem dealing with combinations

$\displaystyle {3 \choose 8}=\frac{8!}{3!(8-3)!}$

3. ## help

Originally Posted by Moo
Hello,

This is a problem dealing with combinations

$\displaystyle {3 \choose 8}=\frac{8!}{3!(8-3)!}$
i know the answer is 56...how to get this answer...

4. You're talking of $\displaystyle \binom{8}{3}$, Moo.

5. Originally Posted by BeBeMala
i know the answer is 56...how to get this answer...

$\displaystyle \frac{{8!}} {{3!\left( {8 - 3} \right)!}} = \frac{{8!}} {{3!5!}} = \frac{{8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2}} {{3 \cdot 2 \cdot 5 \cdot 4 \cdot 3 \cdot 2}} = 8 \cdot 7 = 56$

6. Originally Posted by flyingsquirrel
You're talking of $\displaystyle \binom{8}{3}$, Moo.
the formula to solve this is n!/ (k! (n-k)!),where n=8 and k=3.therefore,when we fill in the numbers we have: 8!/(3!) (8-3)!)
this becomes
40320/(6 *5) =40320/(6*120)=40320/720=56

7. Originally Posted by flyingsquirrel
You're talking of $\displaystyle \binom{8}{3}$, Moo.
I never know which one is above and which one is below