GreenlightTestPrep wrote:
There are n people competing in a chess tournament. Each competitor must play every other competitor k times. If n > 1 and k > 0, what is the total number of games played in the tournament?
A) kn – k
B) (n² – 2k)/2
C) k(n² – n)/2
D) (n² – 2nk + k)/2
E) (kn – 2k)/2
Here's an approach that doesn't require any formal counting techniques:
Let's say a MATCH is when two competitors sit down to play their k games against each other.
If we ask each of the
n competitors, "
How many MATCHES did you have?", the answer will be
n-1, since each competitor plays every other competitor, but does not play against him/herself.
So,
n(n-1) = the total number of MATCHES
IMPORTANT: There's some
duplication here.
For example, when Competitor A says that he/she played n-1 other competitors, this includes the match played against Competitor B. Likewise, when Competitor B says he/she played n-1 other competitors, this includes the match played against Competitor A.
So, in our calculation of
n(n-1) = the total number of MATCHES, we included the A vs B match
twice.
In fact,
we counted every match two times.
To account for this duplication, we'll take
n(n-1) and divide by 2 to get
n(n-1)/2 MATCHES.
Since each match consists of
k games, the total number of games =
kn(n-1)/2Check the answer choices....not there!
However, we can take k
n(n-1)/2 and rewrite it as k(
n² - n)/2
Answer:
Cheers,
Brent