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

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Re: There are 10 children in a company's day-care center, and a pair of [#permalink]
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 day-care center, and a pair of [#permalink]
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 day-care center, and a pair of [#permalink]
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 day-care center, and a pair of [#permalink]
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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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Re: There are 10 children in a company's day-care center, and a pair of [#permalink]
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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 day-care center, and a pair of [#permalink]
Out of 10 children at list a pair means 2

Therefore 10C2 = 10×9/2×1 = 5×9 =45

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Re: There are 10 children in a company's day-care center, and a pair of [#permalink]
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nfa1rhp wrote:
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

Solution:

Since 2 children are selected from 10 and the order of selection doesn’t matter, there are 10C2 = (10 x 9)/2 = 45 different pairs.

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