Question

$4$ gentlemen and $4$ ladies take seats at random round a table. The probability they are sitting alternately is

• A. $\frac{4}{35}$
• B. $\frac{1}{70}$
• C. $\frac{2}{35}$
• D. $\frac{1}{35}$

Answer 1

The correct option is D $\frac{1}{35}$

Let the men take their seats first. They can do so in $(4-1)!=6$ ways.
Now, the ladies would take their seat in the 4 gaps thus created in $4!=24$ ways.
Thus, total ways of seating alternately = $6\ast 24=144$
Thus, probability = $\frac{144}{(8-1)!}=\frac{144}{7!}=\frac{1}{35}$