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!

10 card hand problem

  1. Oct 31, 2015 #1
    1. The problem statement, all variables and given/known data
    How many 10-card hands are there chosen from a standard 52-card deck in which there are exactly two 4-of-a-kinds; no pairs or 3-of-a-kinds?

    3. The attempt at a solution
    if there are exactly 2 4 of a kinds that takes up 8 of the 10 cards in the hand and the remaining 2 must be of different kinds since there are no pairs. first we choose 2 kinds for the 4 of a kinds: 13C2 ways . now we have to make sure the last 2 cards are of different kinds so we pick 2 of the remaining 11 kinds: 11C2 ways. now for each of these kinds there are 4 ways to pick a card from them so 4x4 = 16.

    the whole process is:
    (13C2)(11C2)(16)

    is this correct? I want to make sure I am not double counting anything
     
  2. jcsd
  3. Oct 31, 2015 #2
    Looks good; I got the same answer with the same reasoning before looking at your solution.
     
  4. Oct 31, 2015 #3

    Samy_A

    User Avatar
    Science Advisor
    Homework Helper

    By using a slightly different reasoning, I got ##{{13}\choose{2}}({{44}\choose{2}}-11{{4}\choose{2}})##
    Both our results give 68640, so that looks promising.
     
  5. Oct 31, 2015 #4

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Another way to finish it off:

    Once you have the two fours. The next card can be anything (44 cards) and the last card anything of a different kind (40 cards).
     
  6. Oct 31, 2015 #5
    This doesn't work because you would double count. Order of the draw doesn't matter, so you have to divide by 2!.

    $$16\frac{11!}{2!9!} = \frac{16*11*10}{2} = \frac{44*40}{2}$$
     
  7. Oct 31, 2015 #6

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Yes, that's what I meant!
     
  8. Oct 31, 2015 #7
    thanks everyone
     
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: 10 card hand problem
Loading...