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# Counting problem

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A group of nine people is be split into three teams, so that one team has four people, another team has four people, and the final team has one person. How many ways can this be done?

Apr 16, 2022

#1
+2455
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There are 9 ways to choose a single person.

You then have $$8 \choose 4$$ ways to choose a group of 4 people. But, you have to divide by 2, to correct for overcounting.

Once that is done, you have $${4 \choose 4} = 1$$ way to choose the remaining.

Thus, there are $$9 \times {8 \choose 2} \div 2 = \color{brown}\boxed{315}$$ ways.

All credit goes to Melody: https://web2.0calc.com/questions/a-group-of-nine-people-is-be-split-into-three-teams_1

Apr 16, 2022
#2
+124598
+1

We can choose  any 4 of the 9   people to be on  1 of the 3 teams

And then we can choose any 4 of the 5 remaining people to be on 1 of the remaining 2 teams

The last person is on the last team by default

So

C ( 9,4)  * C ( 3,1) * C(5,4) * C (2,1)  =   3780  ways

Apr 16, 2022
#3
+2455
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