# 4 men,2 women and a child are to be seated around a round table with 7 seats. Find the number of ways they may be arranged if !? a) the child is to be seated between two women b) the child is to be seated between two men

a. 48. b. 288

#### Explanation:

Let's first see that we're working with a table, meaning there is no first seat or last seat (like what we have in a row). What that gives us then is not numbered or distinct seats but instead indistinct seating where we only care about the seating relationships between people.

For a, we're placing the two women on either side of the child. There are 2 ways for the two women to sit.

The four men can sit wherever they'd like in the remaining seats, which is 4! = 24.

All together then, we have

$2 \times 24 = 48$ ways

For b, we're placing two of the four men around the child. We can find the number of arrangements as either a permutation:

(4!)/(2!)=4xx3=12

or as a combination, seeing that each combination of men has two ways they can be arranged:

$2 \left(\begin{matrix}4 \\ 2\end{matrix}\right) = 2 \left(6\right) = 12$

Either way, we get to 12 ways to arrange the men.

We can arrange the remaining 2 men and 2 women in 4! =24 ways. This all gives:

$12 \times 24 = 288$ ways