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Homework Help: Perms and Combs

  1. Jan 17, 2005 #1
    4 children out of 8 will be selected. But two oldest children can not be both chosen. Total number of combinations = ?

    The mutually exclusive situations are really confusing me.
    I know if they were independent, total n. of combinations would be
    [tex]\frac{8!}{(8-4)!4!}=70[/tex]

    I need to subtract the total number of combinations with one of the two oldest boys. I am at a dead end, even though i feel like i know how to do it. My mind kinda went blank.

    PS. This is the last question i need to answer to finish my "beutifully done" homework. help :P I have high expectations from this piece of art. :biggrin:
     
    Last edited: Jan 17, 2005
  2. jcsd
  3. Jan 17, 2005 #2

    Gokul43201

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    You need to subtract the no. of combinations involving both of the oldest boys. How many such combinations will there be ?

    Simple. First two pick both of the oldest boys. This can be done in exactly 1 way. Now you have to pick 2 more boys from the remaining 6. You know how to do this.

    Subtract from original to get final answer.
     
  4. Jan 17, 2005 #3
    Ahhh that would be [tex]^6C_2=15[/tex] so 55 is the final answer. Thx a lot. :smile:

    I have been looking at the question from a different angle since the beginning.
     
    Last edited: Jan 17, 2005
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