Answer to Question #281384 in Discrete Mathematics for yitbarek

Question

) How many ways are there for eight men and five women to stand in a line so that no two women stand next to each other?

Answer 1

There will be 9 places for the 5 women to occupy. They can do it in 9C5 ways.

Moreover, they can arrange themselves in 5! ways.

Also, the men can be arranged in 8! ways.

So, total no of ways = 9C5*8!*5! = 609638400 ways.

or

How many possible ways are there to arrange eight men in a row? It will be 

8P8

=8!=40320

8P8=8!=40320

Now as no two women stand next to each other, we can imagine the situation as

                   * M * M * M * M * M * M * M * M *

Hence, we need to find how many ways we can arrange 5

5 women in the 9

9 possible (as shown above) places,this is actually 

9P5

=9∗8∗7∗6∗5=15120

9P5=9∗8∗7∗6∗5=15120

Now applying the fundamental law of counting (precisely product rule), total number of possible arrangements satisfying both constraints is: 15120∗40320=609638400

15120∗40320=609638400 which is your required/desired answer.

Ans=609638400