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?
Choices:
A) 10  B) 20  C) 40  D) 60  E) 120
Method 1:   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= 
<gadget id="calculator">4+3+2+1</gadget>
<output>10</output>
10   Method 2:   Each person is one participant of 4 games. So there are in all 4*5= 
<gadget id="calculator">4*5</gadget>
<output>20</output>
20 instances of one participant games. But each game has 2 participants so total number of games= 20/2= 
<gadget id="calculator">20/2</gadget>
<output>10</output>
10   10nswer: 10.
<result>A</result>
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?