GMATPrepNow 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

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