how many 3 person committees can be formed in a club with 8 members?
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Hello, This is a problem dealing with combinations $\displaystyle {3 \choose 8}=\frac{8!}{3!(8-3)!}$
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...
You're talking of $\displaystyle \binom{8}{3} $, Moo.
Originally Posted by BeBeMala i know the answer is 56...how to get this answer... $\displaystyle \frac{{8!}} {{3!\left( {8 - 3} \right)!}} = \frac{{8!}} {{3!5!}} = \frac{{8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2}} {{3 \cdot 2 \cdot 5 \cdot 4 \cdot 3 \cdot 2}} = 8 \cdot 7 = 56$
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
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
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