In a tournament with 15 chess players, if each player plays two games

Question

In a tournament with 15 chess players, if each player plays two games against every opponent, how many total games will be played?

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

M11-29

Bunuel · Accepted Answer

In a tournament with 15 chess players, if each player plays two games against every opponent, how many total games will be played?

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

The number of unique pairs that can be formed from 15 players is given by \(C^2_{15} = 105\). As each pair competes against each other twice, the total number of games played amounts to \(2 *105 = 210\).

Answer: C.

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Hope it helps.

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priyankurml · Answer

I used simple formula:  , where n=15; If they play single match, devide the result by 2:  /2.

C 210

alwaysursgd · Answer

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.

vonshuriz · Answer

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

cabelk · Answer

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

rsaraiya · Answer

C

15! / (15-2)!
= 210

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

Thanks.

Bunuel · Answer

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.

jlgdr · Answer

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

mankul · Answer

Please help, How to solve in the above way if the players play 3 matches each?

Bunuel · Answer

In this case the answer would be C^2_{15}*3=315 . Similar questions to practice: how-many-diagonals-does-a-polygon-with-21-sides-have-if-one-101540.html if-10-persons-meet-at-a-reunion-and-each-person-shakes-hands-110622.html 10-business-executives-and-7-chairmen-meet-at-a-conference-126163.html how-many-different-handshakes-are-possible-if-six-girls-129992.html there-are-5-chess-amateurs-playing-in-villa-s-chess-club-127235.html if-each-participant-of-a-chess-tournament-plays-exactly-one-142222.html there-are-8-teams-in-a-certain-league-and-each-team-plays-134582.html there-are-8-teams-in-a-certain-league-and-each-team-plays-132366.html there-are-in-a-certain-league-20-teams-and-each-team-face-107442.html Hope it helps.

candy1221 · Answer

Choosing C because - this looks like a handshake question to me. So formula would be (n*n-1)/2

(15*14)/2

105.

Because each player plays two games(shakes hand twice), 105*2.

Is this line of reasoning correct?

bumpbot · Answer

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In a tournament with 15 chess players, if each player plays two games

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