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There are 10 children in a company's day-care center, and a pair of [#permalink]

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12 Nov 2008, 05:36

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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?

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

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.

Re: There are 10 children in a company's day-care center, and a pair of [#permalink]

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04 Mar 2015, 05:00

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?

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 day-care center, and a pair of [#permalink]

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04 Mar 2015, 05:09

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?

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.
_________________

Re: There are 10 children in a company's day-care center, and a pair of [#permalink]

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01 Apr 2015, 02:38

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.

Re: There are 10 children in a company's day-care center, and a pair of [#permalink]

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13 Dec 2017, 17:30

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