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If each participant of a chess tournament plays exactly one

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If each participant of a chess tournament plays exactly one [#permalink] New post 10 Nov 2012, 13:06
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If each participant of a chess tournament plays exactly one game with each of the remaining participants, then 153 games will be played during the tournament. Find the number of participants.

A. 15
B. 16
C. 17
D. 18
E. 19
[Reveal] Spoiler: OA
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Re: If each participant of a chess tournament plays exactly one [#permalink] New post 10 Nov 2012, 18:00
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derekgmat wrote:
If each participant of a chess tournament plays exactly one game with each of the remaining participants, then 153 games will be played during the tournament. Find the number of participants.

A. 15
B. 16
C. 17
D. 18
E. 19


If the number of participants is 3 (say A, B, C) the number of games played will be 2 (A plays against B and C) + 1 (B plays against C) = 3
Using the same logic, if the number of participants is n, the number of games played will be (n-1) + (n - 2) + (n - 3) + ... 3 + 2 + 1

Given that this sum = 153 = 1 + 2 + 3 + ... ( n - 1)

Sum of first m positive integers is given by m(m+1)/2. So sum of first (n-1) positive integers is (n-1)*n/2

153 = (n-1)*n/2
(n-1)*n = 306
17*18 = 306 (We know that 15^2 = 225 so the two consecutive numbers must be greater than 15. Also, 20^2 = 400 so the two numbers must be less than 20. The pair of numbers in between 15 and 20 whose product ends with 6 is 17 and 18)

So n = 18

Answer (D)
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Re: If each participant of a chess tournament plays exactly one [#permalink] New post 11 Nov 2012, 00:29
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Thanks Karishma

I must admit I am struggling to work through your solution. I looked at the question again and answered it with the approach of working through each of the choices long-hand.

If A, then 15 players means they each will play 14 games (so 14*15), divide the result by two for double counted games:
14*15 = 210
201/2 = 105, not 153, next

If B, 15*16 = 240
240/2 = 120, not 153, next

If C, 16*17 = 274
274/2 = 137, not 153, next

If D, 17*18 = 306
306/2 = 153 ANSWER D

Am I missing any important concepts by answering the question with this long-hand method as opposed to your more structured approach?

Kind regards

Derek
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Re: If each participant of a chess tournament plays exactly one [#permalink] New post 11 Nov 2012, 04:07
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derekgmat wrote:
Thanks Karishma

I must admit I am struggling to work through your solution. I looked at the question again and answered it with the approach of working through each of the choices long-hand.

If A, then 15 players means they each will play 14 games (so 14*15), divide the result by two for double counted games:
14*15 = 210
201/2 = 105, not 153, next

If B, 15*16 = 240
240/2 = 120, not 153, next

If C, 16*17 = 274
274/2 = 137, not 153, next

If D, 17*18 = 306
306/2 = 153 ANSWER D

Am I missing any important concepts by answering the question with this long-hand method as opposed to your more structured approach?

Kind regards

Derek


both are actually doing the same thing. You approach is a logical one while Karishma's is a structured one. But eventually both are doing same thing:
n*(n-1)/2 = 153 ( or 17*18/2 =153)
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Re: If each participant of a chess tournament plays exactly one [#permalink] New post 11 Nov 2012, 04:40
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derekgmat wrote:
If each participant of a chess tournament plays exactly one game with each of the remaining participants, then 153 games will be played during the tournament. Find the number of participants.

A. 15
B. 16
C. 17
D. 18
E. 19


We are basically told that we can choose 153 groups of two players out of n players, thus C^2_n=153 --> \frac{n!}{(n-2)!2!}=153 --> \frac{(n-1)n}{2}=153 --> (n-1)n=306 --> n=18.

Answer: D.

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Hope it helps.
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Re: If each participant of a chess tournament plays exactly one [#permalink] New post 11 Nov 2012, 18:40
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derekgmat wrote:
Thanks Karishma

I must admit I am struggling to work through your solution. I looked at the question again and answered it with the approach of working through each of the choices long-hand.

If A, then 15 players means they each will play 14 games (so 14*15), divide the result by two for double counted games:
14*15 = 210
201/2 = 105, not 153, next

If B, 15*16 = 240
240/2 = 120, not 153, next

If C, 16*17 = 274
274/2 = 137, not 153, next

If D, 17*18 = 306
306/2 = 153 ANSWER D

Am I missing any important concepts by answering the question with this long-hand method as opposed to your more structured approach?

Kind regards

Derek


Your method is absolutely fine. Your logic of multiplying n players by (n-1) games and dividing by 2 is great. The issue with your approach is that you need to calculate for every option. If the numbers were a little larger, you would end up doing calculations a number of times.
For each option, you are doing this calculation (n-1)*n/2
(A) 14*15/2
(B) 15*16/2
etc
You are trying to find the option that will give you the result 153. I have done the same thing. I get (n-1)*n/2 = 153.
Instead of trying all options, you should multiply 153 by 2 to get 306 and then see which two numbers will end in 6. That way you will save a lot of calculations.

14*15 - No
15*16 - No
16*17 - No
17*18 - Yes
18*19 - No

As for the approach used by me to arrive at n*(n-1)/2:
Think of it this way - you make all participants stand in a straight line. The first one comes up and play a game with everyone else i.e. (n-1) games and goes away. The next one comes up and plays a game with all remaining people i.e. (n-2) people and goes away too. This goes on till last two people are left and one comes up, play a game against the other and they both go away. Hence, they end up playing
(n-1) + (n - 2) + .... 3 + 2 + 1 games (total number of games)

You must learn that sum of first m positive integers is given by m(m+1)/2 (very useful to know this)
We need to sum first (n-1) numbers so their sum will be (n-1)n/2

Bunuel has used the combinatorics approach to arrive at n(n-1)/2. There are n people and you want to select as many distinct two people teams as you can (since each person can play against the other only once)
In how many ways can you do that? nC2 ways
nC2 = n(n-1)/2

So the calculation involved is the same in every case. The method of arriving at the equation is different.
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Re: If each participant of a chess tournament plays exactly one [#permalink] New post 28 Dec 2012, 05:32
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derekgmat wrote:
If each participant of a chess tournament plays exactly one game with each of the remaining participants, then 153 games will be played during the tournament. Find the number of participants.

A. 15
B. 16
C. 17
D. 18
E. 19


\frac{P!}{2!(P-2)!}=153
\frac{P * (P-1) * (P-2)!}{(P-2)!}=2*9*17
P*(P-1) = 18*17
P = 18

Answer: D
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Re: If each participant of a chess tournament plays exactly one [#permalink] New post 29 Dec 2012, 02:59
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derekgmat wrote:
If each participant of a chess tournament plays exactly one game with each of the remaining participants, then 153 games will be played during the tournament. Find the number of participants.

A. 15
B. 16
C. 17
D. 18
E. 19



Num of games - 153, The total number of players that can be chosen as a pair would be nC2.

Now look through the options & start form the middle number as the options are in ascending order - 17C2 = 17*16/2 = 6 as last digit (17 *8) ignore
Next number 18c2 = 18*17/2 = 3 as last digit this is the answer

Choice D
Re: If each participant of a chess tournament plays exactly one   [#permalink] 29 Dec 2012, 02:59
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