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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
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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Loser66
 one year ago
Best ResponseYou've already chosen the best response.0I know in this case, 6men are consider is 1, but don't know how to argue further.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0the 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??

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.115 choose 6 is the arrangement that men seat to each other consecutively

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0But the sample space is tooooo large. How to determine it?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0I 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!!!

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.1those are the arrangements the sample space is all possible arrangement regardless of who seats where

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0hey, kid, 15C6 is not one of the options.

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.115C6 is just the men seat consecutively to each other you need to divide that by all possible arrangements to find the probability

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.1i'm still thinking of the sample space!

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.1all possible permutations 15!?

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.1eh let's so 15! was correct possible sample

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.1so 15C6 was not the needed arragement that men seat to each otherr

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0A) \(\dfrac{1}{\left(\begin{matrix}15\\6\end{matrix}\right)}\) B)\(\dfrac{6!}{\left(\begin{matrix}15\\6\end{matrix}\right)}\) C)\(\dfrac{10!}{15!}\) D) \(\dfrac{6!9!}{14!}\) E)\(\dfrac{6!10!}{15!}\)

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.1eh you had a good idea in the start think of men as being one string so we have 10 element to permute 10!

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.1then what we do is permute the men there are 6 of them so 6! the so then 6!10! to permute men as desired!

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.1the error i did in the start is that i permuted a string of men and 9 women without permuting the men as well

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.1E is the answer

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.1do you get it?

kropot72
 one year ago
Best ResponseYou've already chosen the best response.1The 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: \[\large P(6\ consecutive\ men)=\frac{10}{15C6}=\frac{10\times9! \times6!}{15!}=\frac{10!6!}{15!}\]

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Thanks you all. I got it.
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