1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Binomial Coefficient - Factorials Part III

  1. Jul 25, 2013 #1

    reenmachine

    User Avatar
    Gold Member

    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. jcsd
  3. Jul 25, 2013 #2

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Looks correct to me.
     
  4. Jul 25, 2013 #3

    reenmachine

    User Avatar
    Gold Member

    thank you!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Binomial Coefficient - Factorials Part III
  1. Binomial Coefficient (Replies: 2)

  2. Binomial Coefficients (Replies: 6)

  3. Binomial coefficients (Replies: 2)

Loading...