Loser66
  • Loser66
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
Mathematics
katieb
  • katieb
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

Loser66
  • Loser66
I know in this case, 6men are consider is 1, but don't know how to argue further.
Loser66
  • Loser66
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??
xapproachesinfinity
  • xapproachesinfinity
15 choose 6 is the arrangement that men seat to each other consecutively

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

Loser66
  • Loser66
But the sample space is tooooo large. How to determine it?
Loser66
  • Loser66
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!!!
xapproachesinfinity
  • xapproachesinfinity
those are the arrangements the sample space is all possible arrangement regardless of who seats where
Loser66
  • Loser66
hey, kid, 15C6 is not one of the options.
xapproachesinfinity
  • xapproachesinfinity
15C6 is just the men seat consecutively to each other you need to divide that by all possible arrangements to find the probability
Loser66
  • Loser66
how?
xapproachesinfinity
  • xapproachesinfinity
i'm still thinking of the sample space!
xapproachesinfinity
  • xapproachesinfinity
all possible permutations 15!?
xapproachesinfinity
  • xapproachesinfinity
eh let's so 15! was correct possible sample
xapproachesinfinity
  • xapproachesinfinity
so 15C6 was not the needed arragement that men seat to each otherr
Loser66
  • Loser66
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)\(\dfrac{6!10!}{15!}\)
xapproachesinfinity
  • xapproachesinfinity
eh you had a good idea in the start think of men as being one string so we have 10 element to permute 10!
xapproachesinfinity
  • xapproachesinfinity
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!
xapproachesinfinity
  • xapproachesinfinity
the error i did in the start is that i permuted a string of men and 9 women without permuting the men as well
xapproachesinfinity
  • xapproachesinfinity
E is the answer
xapproachesinfinity
  • xapproachesinfinity
do you get it?
kropot72
  • kropot72
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: \[\large P(6\ consecutive\ men)=\frac{10}{15C6}=\frac{10\times9! \times6!}{15!}=\frac{10!6!}{15!}\]
Loser66
  • Loser66
Thanks you all. I got it.

Looking for something else?

Not the answer you are looking for? Search for more explanations.