 Hi children and how are you? I'm Priyanka and let us discuss the following question. It is required to seat 5 men and 4 women in a row so that the women occupy the even place. How many such arrangements are possible? Here we are given 5 men and 4 women. Now representing place numbers of a rows as follows. We have 1, 2, 3, 4, 5, 6, 7, 8, 9. Now we see that even places are 2, 4, 6 and 8. That is 4 places. So women can occupy only these 4 places. Now on using theorem 1 of permutation there will be as many number of ways of seating 4 women as permutation of 4 different places taken 4 at a time. So required number of permutation for women is equal to 4p4. It will be the first equation. Whereas 5 men will occupy the 5 odd places. So required number of permutation is equal to 5e5. That is the second equation and that is 4 men. Applying the multiplication principle we get the total number of arrangements of seating 5 men and 4 women in a row so that women occupies the even number or even places is equal to equation 1 multiplied by equation 2. That is 4p4 multiplied by 5p5. That is 4 factorial divided by 4 minus 4 factorial multiplied by 5 factorial 5 minus 5 factorial. That is equal to 4 factorial into 5 factorial. That can be written as 4 into 3 into 2 into 1, 5 into 4 into 3 into 2 into 1. That is 24 into 120, 2880. And the required answer is 2880. So this completes, I hope you know the term one of permutations. Take care.