10 people are trying to do a tournament of chess. They want to play to games each to all of people. How many games must be played?
+128579 Nice answer, geno.....
Another way to see this is that the 10th person plays 9 other people, the 9th person plays 8 other people (we've already counted the game against he/she played against the 10th person), the 8th person plays 7 other people, etc. Note that the 2nd person only has the 1st person left to play !!!!
So the sum of the first nine integers is just [9 * 10 ] / 2 = 45 games.
+23246 Each of the 10 will have to play 9 games each.
That would be 10 x 9 = 90 games; but ...
Each game would be counted twice: the game that Alan plays against Barb would be one of Alan's 9 but it would also be counted as one of Barb's 9 games.
Therefore, we have to divide 90 by 2 to get a total of 45 games.
+128579 Nice answer, geno.....
Another way to see this is that the 10th person plays 9 other people, the 9th person plays 8 other people (we've already counted the game against he/she played against the 10th person), the 8th person plays 7 other people, etc. Note that the 2nd person only has the 1st person left to play !!!!
So the sum of the first nine integers is just [9 * 10 ] / 2 = 45 games.
