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There are 10 children in a company's day-care center, and a pair of
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Updated on: 04 Mar 2015, 02:59
Updated on: 04 Mar 2015, 02:59
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There are 10 children in a company's day-care 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
A) 100
B) 90
C) 50
D) 45
E) 25
Re: There are 10 children in a company's day-care center, and a pair of
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12 Nov 2008, 06:44
12 Nov 2008, 06:44
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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
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 day-care center, and a pair of
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03 Mar 2015, 22:49
03 Mar 2015, 22:49
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Hi All,
This is an example of a fairly-straightforward 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!(N-K)!] 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
This is an example of a fairly-straightforward 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!(N-K)!] 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
