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There are 10 children in a company's daycare center, and a pair of
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Updated on: 04 Mar 2015, 02:59
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There are 10 children in a company's daycare center, and a pair of children is to be selected to play a game. At most, how many different pairs are possible? A) 100 B) 90 C) 50 D) 45 E) 25
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Originally posted by nfa1rhp on 12 Nov 2008, 06:36.
Last edited by Bunuel on 04 Mar 2015, 02:59, edited 2 times in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.




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Re: There are 10 children in a company's daycare center, and a pair of
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03 Mar 2015, 22:49
Hi All, This is an example of a fairlystraightforward Combinations question. It asks how many different pairs of children (meaning "sets of 2") are there that can be selected from a group of 10 children. In these situations, the ORDER of the two children does not matter, so if 'child A' plays 'child B' then that pair is the SAME as when 'child B' plays 'child A.' As such, you are NOT supposed to count that pairing twice. The Combinations Formula removes all of those duplicate pairs. Combinations = N!/[K!(NK)!] Where N is the total number of children and K is the number that you are going to pick. Here, we have N = 10 and K = 2 10!/[2!8!] = (10)(9)/(2)(1) = 45 different pairs of 2 children. Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: There are 10 children in a company's daycare center, and a pair of
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12 Nov 2008, 06:44
First way 1. first child out of 10  10 possibilities 2. second child out of 9  9 possibilities. the total number of pairs = 9*10=90 3. exclude xy yx cases: 90/2=45 Second way N=10C2=10!/(8!2!)=10*9/2=45
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Re: There are 10 children in a company's daycare center, and a pair of
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04 Mar 2015, 06:00
Hello all,
I also did it like this:
10 children, one pair (2 children) selected and 8 children not:
10!/2!*8! = 9 * 10 / 1*2 = 90 / 2 = 45.
However, if we want to use pairs instead of the number of children, could we also say that out of the 5 pairs we are selecting 1 pair? And how would this go using combinations?



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Re: There are 10 children in a company's daycare center, and a pair of
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04 Mar 2015, 06:05
pacifist85 wrote: Hello all,
I also did it like this:
10 children, one pair (2 children) selected and 8 children not:
10!/2!*8! = 9 * 10 / 1*2 = 90 / 2 = 45.
However, if we want to use pairs instead of the number of children, could we also say that out of the 5 pairs we are selecting 1 pair? And how would this go using combinations? hi pacifist, the meaning of question will change if you write out of the 5 pairs we are selecting 1 pair?.. it means each pair is a separate identity and you have to choose one out of it .. the ans will be 5c1=5...
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There are 10 children in a company's daycare center, and a pair of
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04 Mar 2015, 06:09
chetan2u wrote: pacifist85 wrote: Hello all,
I also did it like this:
10 children, one pair (2 children) selected and 8 children not:
10!/2!*8! = 9 * 10 / 1*2 = 90 / 2 = 45.
However, if we want to use pairs instead of the number of children, could we also say that out of the 5 pairs we are selecting 1 pair? And how would this go using combinations? hi pacifist, the meaning of question will change if you write out of the 5 pairs we are selecting 1 pair?.. it means each pair is a separate identity and you have to choose one out of it .. the ans will be 5c1=5... Yes, I also tested it and ended up with 5. But I wnted to see if there would be another approach to correct for it. For example, 5*9=45. But is this logical?



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Re: There are 10 children in a company's daycare center, and a pair of
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04 Mar 2015, 06:19
pacifist85 wrote: chetan2u wrote: pacifist85 wrote: Hello all,
I also did it like this:
10 children, one pair (2 children) selected and 8 children not:
10!/2!*8! = 9 * 10 / 1*2 = 90 / 2 = 45.
However, if we want to use pairs instead of the number of children, could we also say that out of the 5 pairs we are selecting 1 pair? And how would this go using combinations? hi pacifist, the meaning of question will change if you write out of the 5 pairs we are selecting 1 pair?.. it means each pair is a separate identity and you have to choose one out of it .. the ans will be 5c1=5... Yes, I also tested it and ended up with 5. But I wnted to see if there would be another approach to correct for it. For example, 5*9=45. But is this logical? no pacifist, it will not be correct.
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1) Absolute modulus : http://gmatclub.com/forum/absolutemodulusabetterunderstanding210849.html#p1622372 2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html 3) effects of arithmetic operations : https://gmatclub.com/forum/effectsofarithmeticoperationsonfractions269413.html
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Re: There are 10 children in a company's daycare center, and a pair of
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01 Apr 2015, 03:38
Solving tricky questions like this is really difficult until you do not have taken good basic educations and that is why I have decided to send my daughter to Phoenix kindergarten for her better future.



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Re: There are 10 children in a company's daycare center, and a pair of
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