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Homework Help: Combinations problem

  1. May 4, 2008 #1
    1. The problem statement, all variables and given/known data

    From 10 tickets of luck, only 3 are winning. If we buy 6 tickets, how many possibilities we have at least to have 1 winning ticket in our hand?

    2. Relevant equations

    [tex]C_n^k=\frac{n!}{k!(n-k)!}[/tex]

    3. The attempt at a solution

    6 tickets from 10, we can choose on [itex]C_1_0^6[/itex]. There are 4 tickets left. [itex]C_7^3[/tex] is the tickets which are not winning. Is the right answer:


    [tex]C_1_0^6[/tex] - [tex]C_7^3[/tex]

    ?
     
  2. jcsd
  3. May 4, 2008 #2
    Focus on the "at least" part.
     
  4. May 4, 2008 #3
    Sorry, but I cant figure out
     
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