## 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

1. Loser66

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

2. 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??

3. xapproachesinfinity

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

4. Loser66

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

5. 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!!!

6. xapproachesinfinity

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

7. Loser66

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

8. xapproachesinfinity

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

how?

10. xapproachesinfinity

i'm still thinking of the sample space!

11. xapproachesinfinity

all possible permutations 15!?

12. xapproachesinfinity

eh let's so 15! was correct possible sample

13. xapproachesinfinity

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

14. 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)$$\dfrac{6!10!}{15!}$$

15. 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!

16. 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!

17. 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

18. xapproachesinfinity

19. xapproachesinfinity

do you get it?

20. 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: $\large P(6\ consecutive\ men)=\frac{10}{15C6}=\frac{10\times9! \times6!}{15!}=\frac{10!6!}{15!}$

21. Loser66

Thanks you all. I got it.