in Combinatory
277 views
3 votes
3 votes
How many ways are there to split a dozen people into $3$ teams, where one team has $2$ people, and the other two teams have $5$ people each?
in Combinatory
277 views

1 Answer

3 votes
3 votes
Pick any $2$ of the $12$ people to make the $2$ person team, and then any $5$ of the remaining $10$ for the first team of $5,$ and then the remaining $5$ are on the other team of $5;$ this overcounts by a factor of $2$ though, since there is no designated “first” team of $5.$ So the number of possibilities is  $\left(\left(^{12}\text{C}_{2}\right)\left(^{10}\text{C}_{5}\right)\right)/2 = 8316.$

4 Comments

(12C2) for two people distribution

(10C5) for first group

(5C5) for second group of 5 people each.

((12C2)*(10C5)*(5C5))/2!
0
0
we divided by 2 bcz say the case when two teams B and C have 5 and 5 player each is similar to saying C and B have 5 and 5 player each bcz there are three teams and it is not mentioned which team get 2, 5,5 players (i.e. question doesn’t say like team A should have 2 , B should have 5 and C should have 5 instead its says 3 teams can have 2,5,5 players each)
0
0
DOIB PROBLEMS
0
0
Answer:

Related questions