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Loser66

  • one year ago

At a banquet, 9 women and 6men are to be seated in a row of 15 chairs. If the entire seating arrangement is to be chosen at random, what is the probability that all of the men will be seated next to each other in 6consecutive positions? Please, help

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  1. Loser66
    • one year ago
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    I know in this case, 6men are consider is 1, but don't know how to argue further.

  2. Loser66
    • one year ago
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    the arrangement can be 6men, 9women 1w, 6m, 8w 2w, 6m, 7w ......... hence we have exactly 8 ways to arrange the block of men like that., then??

  3. xapproachesinfinity
    • one year ago
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    15 choose 6 is the arrangement that men seat to each other consecutively

  4. Loser66
    • one year ago
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    But the sample space is tooooo large. How to determine it?

  5. Loser66
    • one year ago
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    I meant: w,m,w,m............is one of the sample space m, w, m, w............is another one ww,m, ww, m,ww... is another one w, m, ww, mm,......is another one. ha!!!

  6. xapproachesinfinity
    • one year ago
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    those are the arrangements the sample space is all possible arrangement regardless of who seats where

  7. Loser66
    • one year ago
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    hey, kid, 15C6 is not one of the options.

  8. xapproachesinfinity
    • one year ago
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    15C6 is just the men seat consecutively to each other you need to divide that by all possible arrangements to find the probability

  9. Loser66
    • one year ago
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    how?

  10. xapproachesinfinity
    • one year ago
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    i'm still thinking of the sample space!

  11. xapproachesinfinity
    • one year ago
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    all possible permutations 15!?

  12. xapproachesinfinity
    • one year ago
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    eh let's so 15! was correct possible sample

  13. xapproachesinfinity
    • one year ago
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    so 15C6 was not the needed arragement that men seat to each otherr

  14. Loser66
    • one year ago
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    A) \(\dfrac{1}{\left(\begin{matrix}15\\6\end{matrix}\right)}\) B)\(\dfrac{6!}{\left(\begin{matrix}15\\6\end{matrix}\right)}\) C)\(\dfrac{10!}{15!}\) D) \(\dfrac{6!9!}{14!}\) E)\(\dfrac{6!10!}{15!}\)

  15. xapproachesinfinity
    • one year ago
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    eh you had a good idea in the start think of men as being one string so we have 10 element to permute 10!

  16. xapproachesinfinity
    • one year ago
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    then what we do is permute the men there are 6 of them so 6! the so then 6!10! to permute men as desired!

  17. xapproachesinfinity
    • one year ago
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    the error i did in the start is that i permuted a string of men and 9 women without permuting the men as well

  18. xapproachesinfinity
    • one year ago
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    E is the answer

  19. xapproachesinfinity
    • one year ago
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    do you get it?

  20. kropot72
    • one year ago
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    The sample space for randomly seating the 6 men is 15C6. The number of ways to have 6 consecutive men is 10. Therefore the required probability is given by: \[\large P(6\ consecutive\ men)=\frac{10}{15C6}=\frac{10\times9! \times6!}{15!}=\frac{10!6!}{15!}\]

  21. Loser66
    • one year ago
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    Thanks you all. I got it.

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