A school teacher has 6 girls and 8 boys to arrange for a choir. Determine the number of different ways she can arrange the 14 different children,
a)if the boys should stand behind the girls.
BBBBBBBB
GGGGGG
Answer = (8!)(6!) = (40320)(720) = 29030400 ways
b)in a single row,if the girls must be kept together.
For each of the 6!=720 arrangements of the girls, there are
9 things to arrange, (that is, 8 boys and 1 row of girls)
Answer = (6!)(9!) = (720)(362880) = 261273600
c)in a single row,if no two girls should stand next to each other.
First we stand the boys in 8!=40320 ways
B B B B B B B B
We put a blank for each place a girl could be inserted between two boys
or on one of the ends
_B_B_B_B_B_B_B_B_
Of those 10 places between two boys or on an end, we choose 6.
So there are 10C6 = 210.
For each of those 210 ways to place the 6 girls, they can be arranged in
6!=720 ways.
Answer = 8!∙10C6∙6! = (40320)(210)(720) = 6096384000 ways.
Edwin