Question

$4$ gentlemen and $4$ ladies take seats at random around a table. What is the probability that they are sitting alternately?

Answer 1

Circular arrangement:

We have to arrange four gentlemen and four ladies around a table.

Let us consider gentlemen take the seat first which can be done in $(4-1)!\mathrm{ways}=3!\mathrm{ways}$

$=3\times 2\times 1=6\mathrm{ways}$.

Now, the ladies will be sitting in four gaps available in $=4!\mathrm{ways}$

$=4\times 3\times 2\times 1=24\mathrm{ways}.$

So, the total number of ways of arranging four gentlemen and four ladies alternatively $=6\times 24=144\mathrm{ways}.$

We know, total number of ways of arranging eight people around a round table $=(8-1)!=7!$

$=7\times 6\times 5\times 4\times 3\times 2\times 1=5040$

So, the required probability $=\frac{\mathbf{No}\mathbf{.}\mathbf{}\mathbf{of}\mathbf{}\mathbf{ways}\mathbf{}\mathbf{of}\mathbf{}\mathbf{arranging}\mathbf{}\mathbf{four}\mathbf{}\mathbf{gents}\mathbf{}\mathbf{and}\mathbf{}\mathbf{four}\mathbf{}\mathbf{ladies}\mathbf{}\mathbf{alternately}}{\mathbf{Total}\mathbf{}\mathbf{ways}\mathbf{}\mathbf{of}\mathbf{}\mathbf{arranging}\mathbf{}\mathbf{eight}\mathbf{}\mathbf{people}\mathbf{}\mathbf{around}\mathbf{}\mathbf{the}\mathbf{}\mathbf{round}\mathbf{}\mathbf{table}}$

$=\frac{144}{5040}=\frac{1}{35}$.

Hence, the probability to arrange four gents and four ladies around a table while sitting alternatively $=\frac{1}{35}$.