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# 15 chess players take part in a tournament. Every player

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CEO
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15 chess players take part in a tournament. Every player [#permalink]  21 Nov 2007, 01:40
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15 chess players take part in a tournament. Every player plays twice with each of his opponents. How many games are to be played?

A. 190
B. 200
C. 210
D. 220
E. 225
[Reveal] Spoiler: OA
Manager
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210

It's 2 time (15 choose 2
or
2(15! / 2!(13!))
= 2(15*7)
=210
CEO
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nice. OA is C 210.

This is a variation of this question

http://www.gmatclub.com/forum/t54046
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Re: 11.29 Combinatorics [#permalink]  10 Aug 2008, 04:19
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I used simple formula: [n*(n-1)], where n=15; If they play single match, devide the result by 2: [n*(n-1)]/2.

C 210
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Re: 11.29 Combinatorics [#permalink]  24 Jul 2009, 18:00
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
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Re: 11.29 Combinatorics [#permalink]  25 Jul 2009, 10:29
15C2 x 2 = 210
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Re: 11.29 Combinatorics [#permalink]  24 Sep 2009, 19:19
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Though 2*(15C2) is the correct approcah to do this, but for people like me who find Perm, Comb n Prob a nightmare, an easy approach can be used.

The first guy has to play 2 matches with the rest of 14, so he'll play 28 matches.

Similarly, second guy has to play with the rest of 13 as his 2 games with the first guy are already played. So he plays 26 matches.

This continues like this and the total matches are 28+26+24...+2

28+26+...+2 = 2(14+13+...+1) = 2((14*15)/2) = 14*15 = 210.

Hope, it is clear to people like me and the ones who already used the shortkut approach.
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Re: 11.29 Combinatorics [#permalink]  27 Sep 2009, 01:18
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
Manager
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Re: 11.29 Combinatorics [#permalink]  11 Oct 2009, 07:13
15th player will play 14 games with all the other players

similarly , 14th player will play 13 games and so on..

so total no.of games = 14*15/2 = 105

since each player plays 2 games with every other player total no.of games = 210

I will go with option C
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Re: 11.29 Combinatorics [#permalink]  04 Mar 2010, 02:54
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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
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Re: 11.29 Combinatorics [#permalink]  04 Mar 2010, 08:54

My approach would be
1st player plays 14 games
2nd plays 13 games
.
14th player plays 1 game

total = 14 + 13 + ...... + 1 = (14*15)/2 = 105

Every plays other player two times so multiply by 2 hence 210.
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Re: 11.29 Combinatorics [#permalink]  13 Oct 2010, 22:06
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vonshuriz wrote:
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
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Re: 11.29 Combinatorics [#permalink]  02 May 2011, 06:23
Hi,

logical approach:

qs states that each player plays twice with each opponent,so if there are n people,2n games will be played per player.

15 chess players will play the number of games which will be a multiple of 15(2n) = 30n.

out of the choices available,the only one divisible by 30 is 210,so thats the answer.

Works well with the options here because all others were not divisible by 30.

however,doing it via combinations is a fool-proof approach

Thanks!
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Re: 15 chess players take part in a tournament. Every player [#permalink]  23 Feb 2012, 12:20
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
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Re: 15 chess players take part in a tournament. Every player [#permalink]  23 Feb 2012, 13:12
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bmwhype2 wrote:
15 chess players take part in a tournament. Every player plays twice with each of his opponents. How many games are to be played?

A. 190
B. 200
C. 210
D. 220
E. 225

# of different pairs possible from 15 players is $$C^2_{15}=105$$, since each pair plays twice between each other than total # of games is 2*105=210.

Similar questions to practice:
there-are-5-chess-amateurs-playing-in-villa-s-chess-club-127235.html
if-10-persons-meet-at-a-reunion-and-each-person-shakes-hands-110622.html

Hope it helps.
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Re: 15 chess players take part in a tournament. Every player [#permalink]  23 Feb 2012, 13:52
C

15! / (15-2)!
= 210

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

Thanks.
Math Expert
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Re: 15 chess players take part in a tournament. Every player [#permalink]  23 Feb 2012, 14:16
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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.

Hope it helps.
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Re: 15 chess players take part in a tournament. Every player [#permalink]  23 Feb 2012, 15:08
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.
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Re: 15 chess players take part in a tournament. Every player [#permalink]  11 Sep 2013, 09:44
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Re: 15 chess players take part in a tournament. Every player [#permalink]  08 Jan 2014, 14:40
bmwhype2 wrote:
15 chess players take part in a tournament. Every player plays twice with each of his opponents. How many games are to be played?

A. 190
B. 200
C. 210
D. 220
E. 225

15*15 - 15 = 210

Normally one would want to divide by 2, but since they are playing against each other twice, it stands as 210

C

Cheers
J
Re: 15 chess players take part in a tournament. Every player   [#permalink] 08 Jan 2014, 14:40

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