$6$ women and $4$ men wait in line. If their order in line is random, find the probability that all of the women are adjacent to one another.

Question

My thoughts on the problem are that the number of ways the women can be adjacent to each other is $5!$ and the total number of arrangements for all the people is $10!$. Is this correct?

Answer 1

Some hints: $10$ people can be arranged in a line in $10!$ ways. Four men and a bench can be arranged in $5!$ ways. Six women can be placed on the bench in $6!$ ways.

Answer 2

There are $10!$ arrangements in total. The number of arrangements where women are adjacent is that $5!\times 6!$. So the probability you are looking for is the ratio of these two.

Answer 3

Total number of arrangements was correctly determined as $10!$.

The number of ways the $6$ women can all stand next to each other are easily seen through the following:

$W_{1} W_{2} W_{3} W_{4} W_{5} W_{6}$ __ __ __ __

__ $W_{1} W_{2} W_{3} W_{4} W_{5} W_{6}$ __ __ __

__ __ $W_{1} W_{2} W_{3} W_{4} W_{5} W_{6}$ __ __

__ __ __ $W_{1} W_{2} W_{3} W_{4} W_{5} W_{6}$ __

__ __ __ __$W_{1} W_{2} W_{3} W_{4} W_{5} W_{6}$

So, we have $5$ ways for the women to be next to each other and $4!$ ways to arrange the men. Next, we need to determine how many different ways the women can be arranged, which is $6!$

Thus, the total number of ways the women can be next to each other is: $(5\cdot 4!)(6!)$

Probability all women are adjacent: $\large\frac{5!6!}{10!}$

Answer 4

The total number of arrangements in which women are next to each other is $5!$ times number of ways in which they can be permuted among themselves which is $6!$. So, the probability is $\dfrac{5!6!}{10!}$

Answer 5

Choosing $6$ places out of $10$ can be done in $\binom{10}{6}$ ways.

Choosing $6$ places out of $10$ under the extra condition that they are adjacent can be done in $5$ ways.

This leads to a probability of: $$5\times\binom{10}{6}^{-1}=\frac{5}{210}=\frac{1}{42}$$

Answer 6

((women factorial - men factorial) $\times$ possible permutations)/Total possible arrangements irrespective of order factorial

((6!-4!)$\times$5)/10! = 3600/3628800 = 0.00099206 or a 0.0992% probability

Answer 7

this is $$1- \frac{\text{number of ways in which they are together}}{\text{total possibilities}} = 1-\frac{5!}{10!}$$