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I like this one..I approached it in a very messy way.The 15th. person has to play 2 games with 14 other ppl(14x2),the 14th person has to play 2 games with 13 ppl(13x2).. So,28+26+24+22..=210 _________________

15 chess players take part in a tournament. Every player plays twice with each of his opponents. How many games are to be played?

190 200 210 220 225

Soln: First lets consider tat each player plays against each other player once and find the total number of games. This can be just multiplied by 2 inorder, to find the number of games when each player plays 2 games which each other. Now first considering that each player plays 1 game against each other, The first player will play 14 games against each of the other 14 opponents numbered from 2 - 15 The second player will play 13 games against each of the other 13 opponents numbered from 3 - 15. ( Note that the game between first player and second player is already counted in the earlier statement for first player) The third player players 12 games against each of the other 12 opponents numbered from 4-15 and so on. Thus we have = 14 + 13 + 12 + ... + 1 = (14 * 15)/2

Now since each player plays two games we can multiply above equation by 2 and we have = (14 * 15 * 2)/2 = 14 * 15 = 210

Hey, I like the easy approach with 28 x 26 x 24 etc.

However I am wondering why you used the Combinations formula. In my opinion order does matter here and therefore I took the Permutation formula. 15!/(15-2)! = 210, which is also the right solution.

Hey, I like the easy approach with 28 x 26 x 24 etc.

However I am wondering why you used the Combinations formula. In my opinion order does matter here and therefore I took the Permutation formula. 15!/(15-2)! = 210, which is also the right solution.

vonshuriz

Order does not matter in this situation. For example, if I play a game of chess against you, that is the same as saying you played a game of chess against me. There are 15 players, so pick any 2 to play in a match: 15C2. Now make them play twice: 15C2 * 2.

The Permutation formula worked by luck because everyone played each opponent twice. Had the question asked for everyone to play 3 games, the Permutation formula would not work unless you accounted for double counting: (15P2 / 2!) * 3 = (15C2) * 3

Re: 15 chess players take part in a tournament. Every player [#permalink]

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23 Feb 2012, 13:20

Here's how I approached it:

- If there were just 5 people... a b c d e playing ONCE against each other then total number of games would be: = 4 (ab, ac, ad, ae) + 3 ( bc, bd, be) + 2 (cd, ce) + 1 (de) i.e. 4+3+2+1 = 10 i.e. Sum of numbers 1 to n-1

Since n is 15, we need to calculate 1+2+....+14 = average of (14,1) * count of numbers between (14,1) = 7.5 * 14

Since they all played TWO Games, FINAL ANSWER = 7.5 * 14 * 2 = 210

Re: 15 chess players take part in a tournament. Every player [#permalink]

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23 Feb 2012, 15:16

1

This post received KUDOS

Expert's post

rsaraiya wrote:

C

15! / (15-2)! = 210

Initially, I wasnt sure if order matters... Are then any alternate approaches?

Thanks.

Order has nothing to do with it. The point is that, every person should play twice with each of his opponents and since there are \(C^2_{15}=105\) different pairs of players possible then each pair to play twice 105*2=210 games are to be played.

Check the links in my previous post for similar problems.

Re: 15 chess players take part in a tournament. Every player [#permalink]

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23 Feb 2012, 16:08

Bunuel wrote:

rsaraiya wrote:

C

15! / (15-2)! = 210

Initially, I wasnt sure if order matters... Are then any alternate approaches?

Thanks.

Order has nothing to do with it. The point is that, every person should play twice with each of his opponents and since there are \(C^2_{15}=105\) different pairs of players possible then each pair to play twice 105*2=210 games are to be played.

Check the links in my previous post for similar problems.

Hope it helps.

Thanks. I guess I wasn't looking at this correctly... Prob & Comb always throw me for a loop.

Re: 15 chess players take part in a tournament. Every player [#permalink]

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11 Sep 2013, 10:44

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