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

*kudos for all correct solutions
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• If each player plays with each other only once then the total number games is simply given by

o \(^nC_2 = \frac{n!}{[2!(n-2)!]} = \frac{[n(n-1)]}{2}\)
o Where n is the number of players and we choose 2 players at a time to play a single game.
• But then each player is playing its competitor “k” times.

o Hence the total number of games would be multiplied “k” times.
o Thus the total number of games = k * Total number of games played once \(= k*n*\frac{(n-1)}{2} = k*\frac{(n^2 – n)}{6}\)
• Hence the correct answer is Option C.

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I will do this logically

If I choose 3 teams to meet only 1 round. then total number of games =3

n= 3 & k=1 & total =3

Plug in the answer choices

A) kn – k ..................................3-1=2.............................Eliminate

B) (n² – 2k)/2......................... (9-2)/2.............................Eliminate

C) k(n² – n)/2...........................1 * (9-3)/2=3....................Keep

D) (n² – 2nk + k)/2..................(9-6+1)/2.........................Eliminate

E) (kn – 2k)/2..........................(3-2)/2.............................Eliminate


Answer: C

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)/2

Check the answer choices....not there!
However, we can take kn(n-1)/2 and rewrite it as k(n² - n)/2

Answer:

Cheers,
Brent
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