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Thread: Perms & Combs Halp!

  1. #1
    Dec 2010

    Perms & Combs Halp!

    20. Find how many ways two numbers can be selected from the numbers 1, 2, ... , 8, 9 so that their sum is:

    (c) divisible by 3, (d) divisible by 5, (e) divisible by 6.

    22. Nine players are to be divided into two teams of four and one umpire.

    (a) In how many ways can the teams be formed?
    (b) If two particular people cannot be on the same team, how many different combinations are possible?

    23. Four adults are standing in a room that has five exits. Each adult is equally likely to leave the room through any one of the five exits.

    (d) What is the probability that no more than two adults come out the same exit?

    There will be more to come :-(
    Last edited by mr fantastic; Dec 5th 2010 at 02:39 AM.
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  2. #2
    MHF Contributor

    Aug 2006
    Hello oexplooitable and welcome to MathHelpForum.
    You should understand that this is not a homework service nor is it a tutorial service. Please either post some of your own work on this problem or explain what you do not understand about the question.

    To get you started. In #20 just make a list of pairs that add to a multiple of five. Then count them.

    In the next one, there are $\displaystyle \dfrac{8!}{(4!)^2\cdot 2}$ ways to divide eight people into two teams of four.
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