# Binomial Coefficient - Factorials Part III

1. Jul 25, 2013

### reenmachine

1. The problem statement, all variables and given/known data
$| \ X \in \mathcal P(\{0,1,2,3,4,5,6,7,8,9\}) : |X|= 4 \ | = \ \ ???$

2. Relevant equations

There's no wording in the exercise , just what I wrote above.If I understood correctly , they asked me to find the cardinality of the set of all subsets of {0,1,2,3,4,5,6,7,8,9} that contains 4 elements.

So $\binom{10}{4} = \frac{10!}{4!6!} = \frac{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6!}{4!6!} = \frac{10 \cdot 9 \cdot 8 \cdot 7}{4!} = \frac{5040}{24} = 210$

So there's 210 elements in $\{X \in \mathcal P(\{0,1,2,3,4,5,6,7,8,9\}) : |X|= 4\}$

thought on this???

thank you!!

Last edited: Jul 25, 2013
2. Jul 25, 2013

### LCKurtz

Looks correct to me.

3. Jul 25, 2013

thank you!