There are four girls and six boys in a class. They are arranged randomly at10 seats.
i.How many arrangements are possible where girls will be occupied first four seats? Write three of them.
Arrange the 4 girls in the first 4 seats in 4! ways.
Arrange the 6 boys in the last 6 seats in 6! ways.
4!6! = (4∙3∙2∙1)(6∙5∙4∙3∙2∙1) = 24∙720 = 17280 ways.
Let the 4 girls be G1, G2, G3, G4
Let the 6 boys be B1, B2, B3, B4, B5, B6
Three of them are:
1. G1,G2,G3,G4,B1,B2,B3,B4,B5,B6
2. G4,G2,G3,G1,B5,B6,B3,B2,B1,B4
3. G3,G1,G4,G2,B6,B1,B5,B4,B3,B2
ii.How many ways are there where a particular boy will be occupied the 7thseat?
Seat the particular boy in the 7th seat. Then you only have 5 boys to seat.
Arrange the 4 girls in the first 4 seats in 4! ways.
Arrange the 5 unseated boys in the last 5 still-vacant seats in 5! ways.
4!5! = (4∙3∙2∙1)(5∙4∙3∙2∙1) = 24∙120 = 2880 ways.
Edwin
