Question

Five men, two women and a child sit at a round table. Find the number of ways of arranging the seven people if the child is seated
(a) between these two women.
(b) between two men.

Answer 1

Answer:

Step-by-step explanation:

Answers

imhkp4u

Imhkp4uSamaritan

Answer:

a) 48 ways

b) 288 ways.

Step-by-step explanation:

According to the question there are 4 men and 2 women and a child.

In the first case you have to find out the number of ways in which the child is to be seated between the two women. It means that we have to always place two women and a child in between them. It will have two ways in case the women exchanges their seat. And the remaining four men can sit anywhere that is 4 factorial.

All together we have 2 * 4! = 2 * 24 = 48 ways.

Now, when we have to find the number of way in which the child can be seated between two men. We will proceed in the very similar way we will place a child in between two men and that two men will be selected randomly and the number of arrangement that is permentation will be

\frac{4!}{2!} = 24 / 2 = 12 ways.

Now the remaining two men and 2 women can be again arranged in 4! ways.

So, altogether we have, 4! * 12 = 288 ways.