Eight chairs are numbered 1 to 8. Two women and 3 men wish to occupy one chair each. First the women choose the chairs from amongst the chairs 1 to 4 and then men select from the remaining chairs. Find the total number of possible arrangements.
We know that,
nPr
According to the question,
W1 can occupy chairs marked 1 to 4 in 4 different way.
W2 can occupy 3 chairs marked 1 to 4 in 3 different ways.
So, total no of ways in which women can occupy the chairs,
4P2 = 4!/(4 – 2)!
= (4 × 3 × 2 × 1)/(2 × 1)
4P2 = 12
Now, 6 chairs will be remaining.
M1 can occupy any of the 6 chairs in 6 different ways,
M2 can occupy any of the remaining 5 chairs in 5 different ways
M3 can occupy any of the remaining 4 chairs in 4 different ways.
So, total no of ways in which men can occupy the chairs,
6P3 = 6!/(6 – 3)!
=120
Hence, total number of ways in which men and women can be seated
4P2x6P3 = 120 × 12
=1440