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# Its easy if you know the Shortcut

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Manager
Joined: 18 Jun 2011
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Its easy if you know the Shortcut [#permalink]  19 Jun 2011, 22:27
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(N/A)

Question Stats:

100% (01:37) correct 0% (00:00) wrong based on 4 sessions
There are a total of 19 students in Mr. Schmidt's gym class. Over the course of a badminton unit, each of the 19 students will compete exactly once against every other student in the class. How many total games of badminton will be played?

A)38
B)171
C)190
D)210
E)361
[Reveal] Spoiler: OA
Math Forum Moderator
Joined: 20 Dec 2010
Posts: 2028
Followers: 140

Kudos [?]: 1181 [0], given: 376

Re: Its easy if you know the Shortcut [#permalink]  20 Jun 2011, 01:04
guygmat wrote:
There are a total of 19 students in Mr. Schmidt's gym class. Over the course of a badminton unit, each of the 19 students will compete exactly once against every other student in the class. How many total games of badminton will be played?

A)38
B)171
C)190
D)210
E)361

Selecting distinct pairs from 19 students.

$$C^{19}_2=\frac{19*18}{2}=19*9=190-19=171$$

Ans: B"
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Manager
Status: GMAT BATTLE - WIN OR DIE
Joined: 02 May 2011
Posts: 175
Concentration: General Management, Entrepreneurship
GMAT Date: 12-22-2011
GPA: 3.81
WE: General Management (Hospitality and Tourism)
Followers: 2

Kudos [?]: 44 [0], given: 13

Re: Its easy if you know the Shortcut [#permalink]  20 Jun 2011, 01:18
guygmat wrote:
There are a total of 19 students in Mr. Schmidt's gym class. Over the course of a badminton unit, each of the 19 students will compete exactly once against every other student in the class. How many total games of badminton will be played?

A)38
B)171
C)190
D)210
E)361

here we go!

19*19=361

361-19=342 number of games wich teams played with each other (2 times with each)
342/2=171 number of games which play for 1 game with each other.

Good luck
Senior Manager
Joined: 12 Dec 2010
Posts: 282
Concentration: Strategy, General Management
GMAT 1: 680 Q49 V34
GMAT 2: 730 Q49 V41
GPA: 4
WE: Consulting (Other)
Followers: 8

Kudos [?]: 38 [0], given: 23

Re: Its easy if you know the Shortcut [#permalink]  21 Jun 2011, 10:12
I would calculate this way- first Student will have chance to play 18 games (given the condn that one can have only one game with the other student), second will have 17 games, third will have 16 games and so on...

So all the # games would be sum of the series 1+2+ .....+18= (1+18)*18/2= 171 so B it is!
Let me know if there's any disconnect!
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Manager
Joined: 16 Mar 2011
Posts: 177
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Kudos [?]: 31 [0], given: 13

Re: Its easy if you know the Shortcut [#permalink]  21 Jun 2011, 11:22
Aarghh why can't I just think of this! If i read the answer it seems to make sense but I was gonna go for A lol.. Good job guys!
Senior Manager
Joined: 12 Dec 2010
Posts: 282
Concentration: Strategy, General Management
GMAT 1: 680 Q49 V34
GMAT 2: 730 Q49 V41
GPA: 4
WE: Consulting (Other)
Followers: 8

Kudos [?]: 38 [0], given: 23

Re: Its easy if you know the Shortcut [#permalink]  22 Jun 2011, 08:56
^^ In due course of time wisdom comes just hang around !
_________________

My GMAT Journey 540->680->730!

~ When the going gets tough, the Tough gets going!

Director
Joined: 01 Feb 2011
Posts: 760
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Kudos [?]: 74 [0], given: 42

Re: Its easy if you know the Shortcut [#permalink]  22 Jun 2011, 15:58
there are 19 students , if each play with all other members .

then total games can be calculated as 18+17....+1 = 171.

alternate approach is to choose 2 distinct out of 19 = 19c2 = 171

Manager
Joined: 19 Apr 2011
Posts: 111
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Kudos [?]: 3 [0], given: 2

Re: Its easy if you know the Shortcut [#permalink]  23 Jun 2011, 01:51
I used this approach
19 students play against 18 others so 19*18
but each game involves 2 players
so total games is 19*18/2 = 171
Re: Its easy if you know the Shortcut   [#permalink] 23 Jun 2011, 01:51
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# Its easy if you know the Shortcut

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