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A Strange Question

  1. Oct 14, 2004 #1
    Can someone help explain the theory behind the bottom question or give me a hint without giving me the answer. It's a bonus question, but I rather get assistance before copying someone completely.

    I already know Permutations and Combinations and the Fundamental Counting Principal, I'm just uncertain of how to apply them to the question. Also, with the numbers being so large I can't write out all the combinations to try and figure out the question that way either.

    In how many ways can 6 girls sit in 10 chairs if sue and sally must sit together?

    Thanks,
     
  2. jcsd
  3. Oct 14, 2004 #2

    Pyrrhus

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    Well first you can calculate the possibility the other girls will sit, then you know 2 of the must be together, so you can calculate the different places those two girls will pick.

    [tex] _{n}P_{r} = \frac{n!}{(n-r)!} [/tex]

    [tex] _{10}P_{4} = \frac{10!}{(6)!} [/tex]

    [tex] _{2}P_{2} = \frac{2!}{(0)!} [/tex]

    [tex] _{2}P_{2}*_{10}P_{4} = 10080 ways [/tex]
     
  4. Oct 14, 2004 #3
    I think u missed a factor ....

    P(10,4) << number of ways 4 girls can get arranged in 10 seats

    6 seats remain ....
    since sally and sue sit together we can club two seats together and say there are 5 seats of which we need to choose one and we can do that in
    P(5,1)

    and ofcourse sally and sue can be arranged amongst themselves in
    P(2,2)
    -- AI
     
  5. Oct 14, 2004 #4
    Thanks, I'm still a bit lost. 2 P 2 = 2
    This is because you can either have 2 people sit or 1 person sit?

    Would 10 P 4 x 6 P 2 work? Final answer = 151200?

    Hmm, I'll have to look over both your answers some more until I figure it out.

    Sue and Sally are supposed to be intpreted as sitting side by side rather than in the same seat, correct?
     
    Last edited: Oct 14, 2004
  6. Oct 14, 2004 #5

    Pyrrhus

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    It would not work, because 2 girls can pick 6 chairs, but that doesn't mean they will sit together, by the way Tenali is right i missed a factor.
     
  7. Oct 14, 2004 #6

    Pyrrhus

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    Side by side, yes
     
  8. Oct 14, 2004 #7
    Thanks - this question has been stumping me. I'm still unsure of which way to go with it.

    If sue and sally are to be beside each other they I figured that there are 18 combinations that allow this.

    If you use SS12345678 to represent the seats. S-S could also be rearranged to work for Sally, Sue and Sue, Sally?
     
  9. Oct 14, 2004 #8

    tony873004

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    Isn't something missing here?

    If you label the seats 1-10, and they are in a circle, then seats 1 & 10 qualify as next to each other. But if they are in a line, they do not.
     
  10. Oct 14, 2004 #9

    Pyrrhus

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    where does it state they are sitting in a circle?
     
  11. Oct 14, 2004 #10
    They are in a line, we haven't been working with circles.
     
  12. Oct 15, 2004 #11

    tony873004

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    How many seat pairs are there that are next to eachother? Let's list them:

    1) 1-2
    2) 2-3
    3) 3-4
    4) 4-5
    5) 5-6
    6) 6-7
    7) 7-8
    8) 8-9
    9) 9-10

    So there are 9 seat pairs. If Sally and Sue must sit together, but it doesn't matter in what order, they can sit as Sally & Sue, or Sue and Sally. So for each pair of seats, there's two ways they can sit.

    So multiply 9 * 2 for 18 different ways Sally and Sue can sit and still be next to each other.

    Now, for each of the 18 different ways Sally and Sue can sit next to each other, how many possibilities are there for the other girls?

    There are 8 remaining seats for the 4 remaining girls. Girl #1 has 8 possibilities. Once she is seated, Girl #2 has 7 possibilities. Once she is seated, Girl #3 has 6 possibilities. Once she is seated, Girl #4 has 5 possibilities.

    8*7*6*5 = 1680 possible ways for the 4 remaining girls to fill the 8 remaining seats for each of the 18 acceptable pairings of Sue & Sally.

    1680 * 18 = 30240 combinations.
     
  13. Oct 15, 2004 #12
    Thanks everyone. The last answer given was the one my teacher wanted, if anyone was wondering. Thanks again.
     
  14. Oct 17, 2004 #13
    Cyclovenom and I made the same obvious mistake :p
    (i wont iterate what our error was .... which is unnecessary anyways)

    Tony is right with his solution but i still have a small nitpicking to do :D

     
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