Question

Ronnie is scheduling a chess tournament in which each player plays every other player once. He created a table and found that each new player adds more games to the tournament in an arithmetic sequence. Answer this set of project questions regarding the chess games.

table:
# of players | # of games
1 | 0
2 | 1
3 | 3
4 | 6
5 | 10

What would the new explicit rule be, if each player played every other player twice? (Hint: Make a table first.)
a.) an = 2(n^2- n)
b.) an = 2(n^2+ n)
c.) an = n^2+ n
d.) an = n^2 - n

Answer 1

Do you see that if the the new player is the nth player, then n new games are added?

Answer 2

Sorry n -1

Answer 3

So for example, in the last row of the top part, going from 4 players to 5, we have 4 new games. Similarly going from 3 players to 4, we have 3 new games.

Answer 4

So, we would expect that for n players, the number of games is the sum of integers from 1 to n-1.

Answer 5

so what does that mean? :\

Answer 6

Do you know a formula for the sum of integers?
http://en.wikipedia.org/wiki/Summation

Answer 7

a formula for this one?

Answer 8

\[\sum_{k=1}^{n} k = \frac{1}{2} n(n-1)\]

Answer 9

i'm not following <=\ ...

Answer 10

it's in " an = 2(n^2- n)" format, the answer. so how do i get from the equation to that?

Answer 11

according to your question an is the no. of games played when each player faces the other twice...

Now, see that in the first case, when each player faces every other player once,... the no. of games played (if there are 'n' players) is... n(n-1)

but, when each player plays two games with every other player.. then the no. of games is sipmly doubled to give,

an= 2n(n-1) = 2(n^2-n)

Answer 12

@bhaweshwebmaster
@yodawgiheardyolikedawgs
you are right, my mistake
did not read carefully enough

Answer 13

wait, what?
i still dont know the answer :(

Answer 14

oohh, i see, now.. thank you!

Answer 15

*sigh* apparently not. i just submitted my project and it said 2(n²-n) is wrong. thankss....

Answer 16

This is a very standard problem, with a twist, which I overlooked in my post above.
Recall that the sum of integers from 1 to n (or 0 to n, it's the same) is:
\[\sum_{k=1}^{n} k = \frac{1}{2} n\ (n+1)\]

If you look at the pattern in your table, when adding player n, you add n-1 new games. Thus, the total number of games in that part is
\[\sum_{k=1}^{n} k = \frac{1}{2} (n-1)\ n\]
The twist is that "each player played every other player twice." We have to multiply our answer by 2.\[(n-1)\ n = n^2 - n\]