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

Question

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

tusharsadawarte · Accepted Answer

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!.

Let me know your thoughts on this. Reply is appreciated.

Let me know your thoughts on this. Reply is appreciated.

chetan2u · Answer

We first choose 3 teams from 10 => 10C3= \frac{10*9*8}{3!}=120 
Now we have 7 teams to choose 3 from => 7C3= \frac{7*6*5}{3!}=35

Total ways =120*35=4200

A

ansarose · Answer

Could you please elaborate this? 
I am confused because of the second step..how do we get that we have to choose 3 from 7 teams?

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frroldao · Answer

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!

DanAustin · Answer

IMO it is A .4200 
since one can select 10C3 and other can select 7C3

New here. Please give kudos and support

TaqdeesAhmad · Answer

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

Answer: A

CEdward · Answer

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

asiddhu2007 · Answer

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

unraveled · Answer

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

Answer A.

41396302717 · Answer

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

Thank you in advance

Arnobdas09 · Answer

10C3 * 7C3 = 120*35= 4200

Posted from my mobile device

yashasvi2 · Answer

when we select 10c3 we select unique player for team A and when we do 7c3 we select unique players for team b. At last, when we multiply Unique Team Members A with unique Team Members B we will never get a repeating result. eg unique X unique = Unique.

aboug123 · Answer

Still doesnt make sense. By your logic, were saying that game ABC vs DEF is different than DEF vs ABC (were basically designating a team A and team B). but nowhere in the question does it make that distinction?

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

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

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