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Math Help - Combination Question

  1. #1
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    Combination Question

    Hey guys, I'd appreciate any help on this stats question... working would also be appreciated so I can learn how its done

    15 People have been given complementary tickets to 3 shows.

    There are:
    - 6 tickets for show 1,
    - 5 for show 2 and
    - 4 for show 3.

    How many ways can the group of 15 be divided between the 3 shows if 2 people refuse to go to show 1 and one person insists on going to show 3?

    Also, if the tickets end up getting randomly distributed, what is the chance that the 3 people get there wish (i.e. persons 1 and 2 don't go to show 1 and person 3 goes to show 3)
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  2. #2
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    Re: Statistics Combination Question

    Quote Originally Posted by marcusman View Post
    15 People have been given complementary tickets to 3 shows. [/FONT][/COLOR]
    There are:
    - 6 tickets for show 1,
    - 5 for show 2 and
    - 4 for show 3.
    How many ways can the group of 15 be divided between the 3 shows if 2 people refuse to go to show 1 and one person insists on going to show 3?
    Also, if the tickets end up getting randomly distributed, what is the chance that the 3 people get there wish (i.e. persons 1 and 2 don't go to show 1 and person 3 goes to show 3)
    First there are \frac{15!}{(6!)(5!)(4!)} ways to give the tickets out without any restrictions whatsoever.

    Now think this way. Give person three a ticket to show three.
    So we have reduced the problem.
    There are \frac{14!}{(6!)(5!)(3!)} ways to give those remaining tickets out without any restrictions whatsoever.
    But we must remove from that number the cases in which person one or person two got a ticket not wanted. Use inclusion/exclusion:
    2\frac{13!}{(5!)(5!)(3!)}-\frac{12!}{(4!)(5!)(3!)}.
    That is the number of ways in which at least one of the two got a ticket not wanted.
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  3. #3
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    Re: Statistics Combination Question

    Hey thanks for the reply. I get all of it except why you minus the 12! equation from the 13! equation
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  4. #4
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    Re: Statistics Combination Question

    Quote Originally Posted by marcusman View Post
    Hey thanks for the reply. I get all of it except why you minus the 12! equation from the 13! equation
    This all about the inclusion/exclusion principle.

    \|A \| stands for the number of elements in set A.

    Do you know to calculate \|A\cup B\|~?
    Let A be the cases where A receives an unwanted ticket.
    Let B be the cases where B receives an unwanted ticket.
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