There are 5 chess amateurs playing in Villa's chess club tournament. If each chess amateur plays with exactly 4 other amateurs, what is the total number of chess games possible to be played in the tournament?
There are different ways of approaching this question.
Take the first amateur. He plays a game with each of the other four i.e. 4 games.
Now take the second one. He has already played a game with the first one. He plays 3 games with the rest of the 3 amateurs i.e. 3 more games are played.
Now take the third amateur. He has already played a game each with the first and the second amateur. Now he plays 2 games with the remaining 2 amateurs so 2 more games are played.
Now go on to the fourth amateur. He has already played 3 games with the first 3 amateurs. He just needs to play a game with the last one i.e. 1 more game is played.
The last amateur has already played 4 games.
Total no of games = 4+3+2+1 = 10
Each person is one participant of 4 games. So there are in all 4*5 = 20 instances of one participant games. But each game has 2 participants so total number of games = 20/2 = 10
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