Question

A person invites a party of $10$ friends at dinner and places them so that $4$ are on one round table and $6$ on the other round table. The number of ways in which he can arrange the guests is

• A. $\frac{10!}{6!}$
• B. $\frac{10!}{24}$
• C. $\frac{9!}{24}$
• D. none of these

Answer 1

The correct option is B

$\frac{10!}{24}$

Explanation for the correct option:

Step 1: Find the number of ways of selecting $4$ people:

Use the combination method to choose $4$ people from a total of $10$ people.

Ways to select $4$ people$={}^{10}C_6$

Step 2: Find ways to arrange $4$ people and $6$ people on the two tables,

respectively:

Ways to arrange $4$ people at the first table$$\begin{gathered} =(4-1)! \\ =3! \end{gathered}$$

Ways to arrange $6$people at second table$$\begin{gathered} =(6-1)! \\ =5! \end{gathered}$$

Step 3: Find the total ways of arranging guests:

Since all the events are happening simultaneously.

So ways to arrange $$\begin{gathered} ={}^{10}C_6·5!·3! \\ =\frac{10!}{24} \end{gathered}$$

Hence, option (B) is the correct answer.