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# 500 - 600 level question

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Intern
Joined: 27 Sep 2009
Posts: 41
500 - 600 level question [#permalink]

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12 Jun 2010, 05:40
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Difficulty:

(N/A)

Question Stats:

89% (01:00) correct 11% (01:40) wrong based on 99 sessions

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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?

171

220

222

340

222
Intern
Joined: 09 Jun 2010
Posts: 21
Re: 500 - 600 level question [#permalink]

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12 Jun 2010, 11:20
From my understanding, each player hast to play vs every other player in the class except himself:
19*18 = 342

This needs to be divided by 2, because otherwise we would count "play & re-play" (A vs B and B vs A - but we don't need the second game).

We end up with:

342/2 = 171

Hope I didn't mix up anything...

I guess, official/easiest way would be $$\frac{n!}{k!(n-k)!}$$

n! = 19 students in class
k! = we take 2 students (for the game) as our "event"

again - 171 games in total
Manager
Joined: 07 Jun 2010
Posts: 50
Location: United States
Re: 500 - 600 level question [#permalink]

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12 Jun 2010, 22:41
Total students = 19
so 1st student will play against 18 , 2nd play against 17 (since 1st already played with him) and so on

so 18+17+16+......+1 = 19*9 = 171
Manager
Joined: 05 Mar 2010
Posts: 217
Re: 500 - 600 level question [#permalink]

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13 Jun 2010, 00:29
Easy and shortest way

its like selecting 2 students out 19 = 19C2 = 171
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Re: 500 - 600 level question   [#permalink] 13 Jun 2010, 00:29
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