Question

Six men and five women sit around a table. In how many distinct ways can they seat themselves so that no two ladies are together?

Answer 1

Answer:

86400 ways.

Step-by-step explanation:

According  to the question there are 6 men and 5 women in total.

We know that in a round table 6 men can be arranged in (6 - 1)! that is 5! ways.

Now since 6 mens are sitting so there will be 6 spaces in between them and we can arrange 5 women in the 6 spaces as 6C5.

Again the 5 woman can be interchanged in 5 factorial ways therefore the total number of ways:

5!*6*5!=86,400.

Answer 2

Answer:

86400

Step-by-step explanation:

Hi,

Let us place all the 6 men around a table, which can be done in

(6 - 1)! ways

= 5! = 120 ways.

Now, in between these 6 men, there are 6 positions in which

they need to placed in any of the 5 positions , in such a

combination no two ladies will be together. Selecting 5 positions

out of 6 can be done in 6C5 ways = 6 ways.

And all the 5 women should placed in the selected 5 positions

and these 5 women can interchange their places among

themselves which could be done in (5) ! = 5! = 120 ways.

Hence, Total number of ways in which they seat themselves that

no 2 ladies would sit together would be 120*6*120

= 86400 ways

Hope, it helps !