Question:
It is required to seat 5 men and 4 women in a row so that the women occupy the even places. How many such arrangements are possible?
Solution:
5 men and 4 women are to be seated in a row such that the women occupy the even places.
The 5 men can be seated in 5! ways. For each arrangement, the 4 women can be seated only at the cross marked places (so that women occupy the even places).
$\mathrm{M} \times \mathrm{M} \times \mathrm{M} \times \mathrm{M} \times \mathrm{M}$
Therefore, the women can be seated in 4! ways.
Thus, possible number of arrangements = 4! × 5! = 24 × 120 = 2880