Asked by Davin
10 boys and 10 girls are eligible to swim in a mixed relay. In how many ways could the relay team be chosen if the team must contain 2 boys and 2 girls.
Note: the order of the swimmers on the relay matters.
Note: the order of the swimmers on the relay matters.
Answers
Answered by
oobleck
There are 10C2 ways to choose the girls and the boys.
There are 4! ways to arrange the 4 choices. So,
10C2 * 10C2 * 4! = 90^2 * 24 = 194,400 ways
There are 4! ways to arrange the 4 choices. So,
10C2 * 10C2 * 4! = 90^2 * 24 = 194,400 ways
Answered by
Damon
first choose which two girls will race (no order yet)
combinations of 10 girls 2 at a time = 10!/[ 2! (8!) ] = 45
same for the boys 45 ways
for each choice of two girls I can chose a choice of two boys
so if I choose group 1 of boys, I can choose any of the 45 groups of girls
if I choose group 2 of two boys , I can choose any of the 45 groups of girls
etc
45*45 ways
now I have four people in a room, how many ways can I order them
4 * 3 * 2 * 1 =4! = 24
so I end up with 45 * 45 * 24
I do not want to be the coach
combinations of 10 girls 2 at a time = 10!/[ 2! (8!) ] = 45
same for the boys 45 ways
for each choice of two girls I can chose a choice of two boys
so if I choose group 1 of boys, I can choose any of the 45 groups of girls
if I choose group 2 of two boys , I can choose any of the 45 groups of girls
etc
45*45 ways
now I have four people in a room, how many ways can I order them
4 * 3 * 2 * 1 =4! = 24
so I end up with 45 * 45 * 24
I do not want to be the coach