Question

Nine children are to be divided into an A team, a B team and a C team of 3 each. The A team will play in one league, the B team in another, the C team in a third league. How many different divisions are possible

Answer 1

The answer is "840".

Explanation:

Following are the number of ways in which selecting a team A by 9 children:

`= ^{9_{C_(3)}\\\\\\`

`=(9!)/(3! * 6!) \\\\=(9* 8* 7* 6!)/(3 * 2* 1* 6!)\\\\=(9* 8* 7)/(3 * 2* 1)\\\\=(3* 4* 7)/(1)\\\\=(84)/(1)\\\\=84`

Following are the number of ways in which selecting a team B by remaining 6 children:

`= ^(6)_{C_(3)}`

`= (6!)/((3! * 3!))\\\\= (6!)/((3* 2* 1 * 3!))\\\\= (6* 5 * 4 * 3!)/((3* 2* 1 * 3!))\\\\= ( 5 * 4 * 3!)/(3!)\\\\= 5 * 4 \\\\=20`

Following are the number of ways in which selecting a team C by remaining 3 children:

`= ^(3)_{C_(3)}\\\\=(3!)/(3!)\\\\= 1`

Following are the number of ways in which making 3 teams by 9 children:

`= ((84 * 20 * 1))/(3!)\\\\= ((84 * 20 ))/(6)\\\\= 14 * 20\\\\= 280\\\\`

(Note: we've split by 3! Because it also is necessary to implement three teams between themselves)

Now 3 leagues have to be played. One is going to be run by each team.

That is the way it is

Different possible divisions

`= 280 * 3!\\\\= 280 * (3 * 2 * 1)\\\\= 840`