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bmwhype2
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Bunuel
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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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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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vonshuriz
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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C

15! / (15-2)!
= 210

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

Thanks.
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rsaraiya
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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bmwhype2
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 :)
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jlgdr


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 :)

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