In how many ways can nine students be partitioned into three teams containing four, three, and two students respectively?
2007-06-08 22:30:32 · 5 answers · asked by kash 1 in Science & Mathematics ➔ Mathematics
any 4 of 9 students can join the team that has 4 people.
9C4
After 4 students join the team. There are 5 students who can join the team the as 3 people. Any 3 of 5 students can be selected.
5C3
Now there are 2 students who can join the team the has two people. SInce there are two of them left, they have to join no matter what. The combination is 1
the total is 9C4 x 5C3 x 1
2007-06-08 22:37:59 · answer #1 · answered by 7 · 1⤊ 0⤋
from nine select 3. 9C4 then from the remeainin 5 select 3 so 5C3 and then from the remaining 2 select 2 2C2
so no of ways = 9C4 * 5C3 * 2C2
= 9!/4!5! * 5!/3!2! *1
= 9x8x7x6/4x3x2 * 5x4/2
= 126* 10
= 1260 ways
2007-06-09 05:54:01 · answer #2 · answered by Anonymous · 0⤊ 0⤋
This question was already asked by one of your classmates. If you want more input, mine, and another answerers, responces are there. If you are just looking for the answer it is 1260
2007-06-09 06:01:26 · answer #3 · answered by MathIsFun 3 · 0⤊ 0⤋
(9C4)(5C3)(2C2) = 1260
(9C4)(5C2)(3C3) = 1260
(9C3)(6C4)(2C2) = 1260
(9C3)(6C2)(4C4) = 1260
(9C2)(7C4)(3C3) = 1260
(9C2)(7C3)(4C4) = 1260
2007-06-09 05:53:29 · answer #4 · answered by Helmut 7 · 0⤊ 0⤋
9C4*5C3*2C2
2007-06-09 07:30:59 · answer #5 · answered by ? 4 · 0⤊ 0⤋