Given $20$ people – $10$ males and $10$ females that are sitting around a round table.
Find the probability for which the order of the sitting is the following:
Between any pair that consists of $2$ females, there is at least one male.
For example a valid sitting order can be of the form:
$\langle$female,female,male,female,female,male,male,female,female,male,female,female,male …$\rangle$
The answer should be one of the following:
(a) $ 0.0655 $
(b) $ 0.0131 $
(c) $ 0.0027 $
(d) $ 0.0014 $
I tried combining each pair of females into one and finding the number of arrangements around the table to get an upper bound for the answer, but I am getting a number which is lower than the options which available, this is what I did:
$\displaystyle{\frac{\displaystyle{10 \choose 2}2!14!}{19!}}$
There are $\displaystyle{10 \choose 2}$ ways to split the females into groups of $2$, because the order matters here I multiplied it by $2!$, then I multiplied by the number of ways to arrange $15$ people around a round table which is $(15-1)!$ but I am getting a number which is lower than the options for the answer in the question so I'm stuck.
I'll appreciate any insights, thanks !
Best Answer
Hand each man a chair and have the men wait while we seat the women.
In a circular permutation, only the relative order matters. Suppose Anna is one of the females. Seat her. We will use her as our reference point.
There are nine ways to choose another female to sit adjacent to Anna and two ways to choose on which side of Anna that person sits. Once they are seated, there are $8!$ ways to arrange the other eight females as we proceed clockwise around the table from the pair that includes Anna. Doing so determines the other four pairs of females.
Now that we have set the pairs, this creates five gaps in which to place the ten males. Let $x_i$ be the number of males in the $i$th gap as we proceed clockwise around the table from Anna. Then $$x_1 + x_2 + x_3 + x_4 + x_5 = 10 \tag{1}$$ is an equation in the positive integers. A particular solution of Equation 1 corresponds to the placement of four addition signs in the nine spaces between successive ones in a row of ten ones. For instance, $$1 + 1 1 1 + 1 1 + 1 1 1 + 1$$ corresponds to the solution $x_1 = 1, x_2 = 3, x_3 = 2, x_4 = 3, x_5 = 1$. The number of solutions to Equation 1 is the number of ways we can place $5 - 1 = 4$ addition signs in the $10 - 1 = 9$ gaps between successive ones in a row of ten ones, which is $$\binom{10 - 1}{5 - 1} = \binom{9}{4}$$
Now that we have decided how many males are in each gap, there are $10!$ ways to arrange the ten males in those gaps.
Hence, the number of favorable cases is $$9 \cdot 2 \cdot 8! \cdot \binom{9}{4} \cdot 10! = 2 \cdot 9! \binom{9}{4} 10!$$
Since there are $19!$ ways to arrange the other $19$ people as we proceed clockwise around the table from Anna, the probability that there is at least one male between each of the five pairs of adjacent females is $$\frac{2 \cdot 9! \dbinom{9}{4}10!}{19!} \approx 0.0027$$