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

Question

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 · Answer

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

loser66 · Answer

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 · Answer

15 choose 6 is the arrangement that men seat to each other consecutively

loser66 · Answer

But the sample space is tooooo large. How to determine it?

loser66 · Answer

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 · Answer

those are the arrangements 
the sample space is all possible arrangement regardless of who seats where

loser66 · Answer

hey, kid, 15C6 is not one of the options.

xapproachesinfinity · Answer

15C6 is just the men seat consecutively to each other
you need to divide that by all possible arrangements  to find the probability

loser66 · Answer

how?

xapproachesinfinity · Answer

i'm still thinking of  the sample space!

xapproachesinfinity · Answer

all possible permutations 15!?

xapproachesinfinity · Answer

eh let's so 15! was correct possible sample

xapproachesinfinity · Answer

so  15C6 was not the needed arragement that men seat to each otherr

loser66 · Answer

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 · Answer

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 · Answer

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 · Answer

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 · Answer

E is the answer

xapproachesinfinity · Answer

do you get it?

kropot72 · Answer

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 · Answer

Thanks you all. I got it.