Question

The number of ways in which $5$ ladies and $7$ gentlemen can be seated in a round table so that no two ladies sit together, is equal to:

• A. $\frac{7}{2}{\left(720\right)}^2$
• B. $7{\left(360\right)}^2$
• C. $7{\left(720\right)}^2$
• D. $720$
• E. $360$

Answer 1

The correct option is A

$\frac{7}{2}{\left(720\right)}^2$

Explanation for the correct option(s)

Find the number of ways

Consider the given data as,

There are $5$ ladies and $7$ gentlemen to be seated at a round table

Number of ways for gentlemen is Equal to $\left(n-1\right)!$

Then, $7$gentlemen can be seated as,

$$\begin{gathered} \left(7-1\right)!=6! \\ =6\times 5\times 4\times 3\times 2\times 1 \\ =720\text{Ways} \end{gathered}$$

Now for $5$ ladies to be seated at the table in $7$ places that can be expressed as

$$\begin{gathered} {}^7{P}_5=\frac{7!}{\left(7-5\right)!} \\ =\frac{7!}{2!} \\ =\frac{7!}{2\times 1} \\ =\frac{7!}{2} \end{gathered}$$.

Number of arrangements can be expressed as

$$\begin{gathered} 6!\times {}^7{P}_5=720\times \frac{7!}{2} \\ =720\times \frac{7\times 6\times 5\times 4\times 3\times 2\times 1}{2} \\ =720\times \frac{7}{2}\times 720 \\ =\frac{7}{2}{\left(720\right)}^2 \end{gathered}$$

Hence, option A is correct.