Question 568942
each side must have 1 boy and 1 girl.
this makes the number of possible teams equal to 4 * 8 = 32.
each girl can be paired with any of the 8 boys.


it takes 2 teams to play a game.
the games can be arranged in 32 * 31 / 2 = 496 ways.
32 * 31 / 2 is the combination formula of 32C2.


this would be easier to actually show what happens if there were less boys and girls.


assume there were 2 girls and 3 boys
identify the girls by the letter a and b
identify the boys by the numbers 1 and 2 and 3.
there are 2 * 3 = 6 possible teams.
those teams are
a1
a2
a3
b1
b2
b3
it takes 2 teams to play a game.
the number of possible games where each time plays each other team only once would be given by the formula 6C2 which results in 6 * 5 / 2 which results in 15 possible games.
the possible games to be arranged would be as follows:
-----
a1 plays a2
a1 plays a3
a1 plays b1
a1 plays b2
a1 plays b3
-----
a2 plays a3
a2 plays b1
a2 plays b2
a2 plays b3
-----
a3 plays b1
a3 plays b2
a3 plays b3
-----
b1 plays b2
b1 plays b3
-----
b2 plays b3
-----
the formula can be shown to be applicable with 2 girls and 3 boys.
the same formula is used with 4 girls and 8 boys to get the answer provided above.
number of possible teams if 4 * 8 = 32
number of possible games is 32C2 which is equal to 32 * 31 / 2 = 496.