There are 8 teams in a certain league and each team plays each of the other teams exactly once. If each game is played by 2 teams, what is the total number of games played?
A. 15
B. 16
C. 28
D. 56
E. 64
Answer: C
Source: Official Guide
5-Day Free Trial
5-day free, full-access trial TTP
Available with Beat the GMAT members only code
MORE DETAILS
There are 8 teams in a certain league and each team plays each of the other teams exactly once. If each game is played
This topic has expert replies
GMAT/MBA Expert
-
Scott@TargetTestPrep
- GMAT Instructor
- Posts: 7247
- Joined: Sat Apr 25, 2015 10:56 am
- Location: Los Angeles, CA
- Thanked: 43 times
- Followed by:29 members
Solution:
When each team plays each of the other teams exactly once, the total number of games played is 8C2 = 8! / (2! x 6!) = (8 x 7) / 2! = 28 games.
Answer: C
Scott Woodbury-Stewart
Founder and CEO
[email protected]
See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews
One needs to always find \(n\) and \(r\) in permutation and combination problems. \(n\) is the total number from which you choose and \(r\) is how many you choose at a time which makes it one possibility. If the order of the entities chosen at a time, does not matter as in this case, it is \(nCr\), otherwise it is \(nPr\).
\(n\) and \(r\) in this problem are \(8\) and \(2\) respectively and the total number of games played is \(8C2=28\).
Therefore, C
-
deloitte247
- Legendary Member
- Posts: 2214
- Joined: Fri Mar 02, 2018 2:22 pm
- Followed by:5 members
Total no. of teams = 8
Each game is played by 2 teams
Total no. of games played = no. of different pairs from 8 teams
This is going to be selecting 2 from 8 teams which is =>
$$8C_2=>\frac{8!}{2!\left(8-2\right)!}=\frac{8\cdot7\cdot6!}{2\cdot1\cdot6!}=\frac{8\cdot7}{2\cdot1}$$
$$8C_2=>\frac{56}{2}=28$$
Answer = option C
Each game is played by 2 teams
Total no. of games played = no. of different pairs from 8 teams
This is going to be selecting 2 from 8 teams which is =>
$$8C_2=>\frac{8!}{2!\left(8-2\right)!}=\frac{8\cdot7\cdot6!}{2\cdot1\cdot6!}=\frac{8\cdot7}{2\cdot1}$$
$$8C_2=>\frac{56}{2}=28$$
Answer = option C
