At a banquet, $9$ women and $6$ men 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 sated next to each other in $6$ consecutive positions?
(A) $\dfrac1{{15}\choose{6}}$
(B) $\dfrac{6!}{{15}\choose{6}}$
(C) $\dfrac{10!}{15!}$
(D) $\dfrac{6!10!}{15!}$
(E) $\dfrac{6!10!}{15!}$
The correct answer is E, with the explanation as below:
We will use the Fundamental counting principle extensively.
First, the total number of permutations of men and women is $15!$.
To find the number of permutations where all the men are seated next to each other, consider the collection of men to be one unit, which gives us $10!$ permutations of the women and the "men-unit".
Within the "men-unit", there are $6!$ ways to seat the group of men. It follows that there are a total of $10!6!$ ways to seat the group if all the men sit together.
Thus, the probability that all the men sit together is $\dfrac{10!6!}{15!}$
Could anyone explain for me why we divide by $15!$ in the answer?
Note that a probability $P$ is calculated as
$$P = \frac{\text{number of choices that count}}{\text{number of all choices}}$$
$15!$ is the total number of arrangements of $15$ people, so we have to divide by it.