Question

3 ladies and 3 gents can be seated at a round table so that any two and only two of the
ladies sit together. The number of ways is
(a) 70
(b) 27
(c) 72
(d) none of these​

Answer 1

Given:

Number of ladies = 3

Number of gents = 3

To find:

Number of ways such that Any two and only two of the ladies have to sit together.

Solution:

First of all, let us find the arrangements of ladies.

We have 2 places and 3 ladies are there.

Number of ways so that ladies can be sit = $_3P_2 = \frac{3!}{(3-2)!} =6$

Now, it is a circular arrangement and 1 lady is left with 4 possible positions.

This lady can not be adjacent to other ladies, so only 2 possible positions for this lady.

Number of ways to arrange the third lady = $_2P_1 =2$

Number of ways to arrange ladies = 6 $\times$ 2 = 12

Now, there are 3 positions left for 3 gents, so number of ways to arrange the 3 gents:

$_3P_3 = \dfrac{3!}{(3-3)!} = 6$

Therefore, total number of ways to arrange all the persons as per given conditions = 12 $\times$ 6 = 72