Question

5 men and 3 women are seated randomly in a row. Find the probability that the women all sit together.

Answer 1

The solution says treat the people as 8 objects, 5 of one type and 3 of another (5M, 3W). The sample space has \[\frac{8!}{5!3!}=56\]. If we treat the women as a unit "WWW", then there are 6 objects (5M, 1WWW) so \[\frac{6!}{5!1!}=6\]
Hence P = 6/56

I find it odd though that they treat everyone as the same? Wouldn't each person be unique? I tried to reason that if all women sit together, there was 3! ways to do so, and for each of these ways, there are 5! (5x4x3x2x1) ways to seat the men. And the sample space would be 8!, so

P = (3!*5!)/8! = 1/56

Answer 2

Reasoning is quite close. However, instead of 5! in the numerator, it should be 6! because as you mentioned, there are 6 objects to arrange.

Answer 3

Oh really? It wouldn't be overcounting since we count the women in the 3! factor?

Answer 4

Think of it this way. To derive the numerator, we do the ff:
1. First, arrange the 6 objects (5 M's, 1 WWW) in the row. (6!)
2. Then arrange the 3 women in the WWW: (3!)
Got it? :)

Answer 5

Oh tricky. I see now thanks :) (I am awful at combinatorics :( )

Answer 6

Just keep on doing problems like that. I once sucked at combinatorics too. =))

Answer 7

Wait this is strange.. I calculated the result and it gave me 6/56, which is the same as the book tried to do it, counting everyone as "the same".

Answer 8

So I am wondering why treating the people as unique, or the same, gives the same result o_o

Answer 9

Wow. Weird. Well anyway, you don't have to do it the way the book does.

Answer 10

lol ok well thanks

Answer 11

No prob. :)