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# There are 10 people to play in the tournament in which a team of 3 wil

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Re: There are 10 people to play in the tournament in which a team of 3 wil [#permalink]
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IMO A. Followed the same approach as chetan2u

ansana - as the first 3 people (out of the 10) are already chosen, your "new universe" now is only 7. So from those 7, you need to make another team of 3 (i.e., combinations of 7 people grouped in 3)

Hope this helps!
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There are 10 people to play in the tournament in which a team of 3 wil [#permalink]
IMO it is A .4200
since one can select 10C3 and other can select 7C3

New here. Please give kudos and support

Originally posted by DanAustin on 05 Jan 2021, 10:21.
Last edited by DanAustin on 05 Jan 2021, 10:23, edited 1 time in total.
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Re: There are 10 people to play in the tournament in which a team of 3 wil [#permalink]
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ansana wrote:
I am confused because of the second step..how do we get that we have to choose 3 from 7 teams?

Posted from my mobile device

At the start, you have 10 people to choose from to make a team of 3. Once you do make a team, you have now 7 people left to make the other team. So,

$$= 10C3*7C3 = 120 * 35 = 4200$$

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Re: There are 10 people to play in the tournament in which a team of 3 wil [#permalink]
Let's form team 1:

10 x 9 x 8 / 3! = 10 x 3 x 4 = 120 ways <---# of ways team 1 can be formed

7 x 6 x 5 / 3! = 7 x 5 = 35 ways <--- # of ways team 2 can be formed

120 x 35 = 4200 <--- # of ways the teams can be matched up

Ans is A IMO
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Re: There are 10 people to play in the tournament in which a team of 3 wil [#permalink]
I feel it can also be done by:

6 people chosen at first out of 10 in 10C6 ways
Among those 6 people, teams of 3 formed by 6C3 ways
10C6 = 210
6C3=20
210*20 = 4200
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There are 10 people to play in the tournament in which a team of 3 wil [#permalink]
Bunuel wrote:
There are 10 people to play in the tournament in which a team of 3 will play another team of 3 in each game. How many different games can be scheduled in the tournament?

A. 4200
B. 1400
C. 120
D. 35
E. 20

The question doesn't mention that players can't be in more than one team, hence it is possible that one player plays for many teams.
For example, the Player 1 can play for team of 3 consisting player 1,2 and 3 and so for team of 3 consisting player 1, 9 and 10.

Hence there are a lot of possibilities which we need to find.

Ways to choose a team of 3 from 10 = $$10_{C_3} = 120$$
Once the team is selected it would play a match with players from other team that is going to get selected from the remaining 7.

Ways to select a team of 3 from 7 = $$7_{C_3} = 35$$

Then again after the match we have 10 people from which the teams are selected - the above process of selecting teams is repeated and so on.

Hence total number of matches possible = 35*120 = 4200

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Re: There are 10 people to play in the tournament in which a team of 3 wil [#permalink]
Hello
I started this question with the understanding that there are 10 people,
1. divide them into 3 groups
2. then calculate how many games can be there between these 3 groups.

Now, the first part is calculated as .
selecting 9 people from a pool of 10 i.e. 10C9 = 10, now multiplying this with the number of groups that can be made with these 9 people i.e. 9! / (3! * 3! * 3! * 3!). Hence for the first part, the answer comes out to be (10 * 9!) / (3! * 3! * 3! * 3!).

1. 2800

2. calculating number of games 3 teams can play differently : 3

therefore, my answer comes out to be 2800 * 3 = 8400

It would be helpful if somebody could point out the flaw in my reasoning

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Re: There are 10 people to play in the tournament in which a team of 3 wil [#permalink]
10C3 * 7C3 = 120*35= 4200

Posted from my mobile device
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Re: There are 10 people to play in the tournament in which a team of 3 wil [#permalink]
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I think the answer should be 2100. Since, after choosing the first team i.e. 10C3, we are choosing second team 7C3 which is making sense as Team A vs Team B. But if we do 10C3 * 7C3, we are taking in consideration both cases Team A vs Team B and Team B vs Team A (order wise). Hence, it should be divided by 2!.