M23-08

Question

12 teams are set to compete in a competition. If every team faces each of its opponents once, how many games will be played in total?

A. 64
B. 66
C. 82
D. 106
E. 132

Bunuel · Answer

Official Solution:

12 teams are set to compete in a competition. If every team faces each of its opponents once, how many games will be played in total?

A. 64
B. 66
C. 82
D. 106
E. 132

The number of games is equivalent to the total number of  distinct pairs  that can be formed from 12 teams (one game per pair); therefore, the total number of games will be  C^2_{12} = \frac{12!}{10! * 2!} = 66 .

Answer: B

Raihanuddin · Answer

We can do this type of problem by always multiplying the last two numbers and then dividing the product by 2

(12*11)/2 = 66

Bunuel · Answer

Check below: nine-dogs-are-split-into-3-groups-to-pull-one-of-three-88685.html in-how-many-different-ways-can-a-group-of-8-people-be-99053.html 6-people-form-groups-of-2-for-a-practical-work-each-group-95344.html a-group-of-8-friends-want-to-play-doubles-tennis-how-many-55369.html a-group-of-8-friends-want-to-play-doubles-tennis-how-many-106277.html in-how-many-different-ways-can-a-group-of-8-people-be-divide-85707.html how-many-ways-are-there-to-split-a-group-of-6-boys-into-two-105381.html anthony-and-michael-sit-on-the-six-member-board-of-directors-102027.html in-how-many-different-ways-can-a-group-of-9-people-be-divide-101722.html Hope it helps.

sdgmat89 · Answer

why cant we consider (a,b) (b,a) as same here....because when you chose 2 it can be anyway round??.pl suggest

Bunuel · Answer

C^2_{12}  yields different groups of 2 out of 12, when order does not matter. So, you'll get only one group with two different teams, say (a, b). Since (a, b) is the same as (b, a) (a playing with b means b is playing with a), we don't need to consider (b, a) here. So, basically, the order does not matter.

TippingPoint93 · Answer

Hi Bunuel or anyone that could answer this:

Had each team played 3 times instead of once, would the answer be 66*3! Or just 66*3?

+1 for an explanation

Bunuel · Answer

Just 66*3: each pair plays thrice.

KARISHMA315 · Answer

Hi Bunuel , Can we also use the logic - First team will pay with 11 other teams , second team with 10 teams , 3rd team with 9 teams and so on . 
So Total Matches = 11+10+9+8+7+6+5+4+3+2+1 = 66

Bunuel · Answer

Yes, you can do this way too.

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

bumpbot · Answer

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M23-08

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