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# PS - Number Prop

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Senior Manager
Joined: 10 Mar 2008
Posts: 360

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04 Jan 2009, 13:35
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Of the three digit integers greater than 700, how many have two digits that are equal to each other and the remaining digit different from the other two?

A) 90
B) 82
C) 80
D) 45
E)36
SVP
Joined: 29 Aug 2007
Posts: 2473
Re: PS - Number Prop [#permalink]

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04 Jan 2009, 14:01
vksunder wrote:
Of the three digit integers greater than 700, how many have two digits that are equal to each other and the remaining digit different from the other two?

A) 90
B) 82
C) 80
D) 45
E)36

1. xxy position : 3 integers (7, 8 and 9) can fit in place of x and 9 integers, except the one that goes in x's place, from 0-9 can fit in y's place. so the no of ways is = 3x9 = 27

2. xyx position : same here = 27

3. yxx position : 3 integers (7, 8 and 9) can fit in place of y and 9 integers, except the one that goes in y's place, from 0-9 can fit in x's place. so the no of ways is = 3x9 = 27

but take out 700 as it is not > itself.
total = 27+27+27 -1= 80
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Manager
Joined: 15 Apr 2008
Posts: 164
Re: PS - Number Prop [#permalink]

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04 Jan 2009, 14:13
Intern
Joined: 01 Jan 2009
Posts: 30
Re: PS - Number Prop [#permalink]

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04 Jan 2009, 14:44
IMO C

If the number is XYZ

X=Y case = > 1st position can be filled with any of 7,8 or 9 ; so X can be 3 different numbers. Now next positions can be filled out by X , so combinations and the 3rd number can be filled out from any number other than X , so total number of combinations = 3* 1* 9 = 27

Now, there are 2 more similar cases 1 ) X=Z and 2 ) Y=Z ; so total number of combinations = 27* 3 =81

Now, from Y=Z case , we need to eliminate one combination as 700 is excluded from this ; so, total number fo combination= 81-1= 80
Re: PS - Number Prop   [#permalink] 04 Jan 2009, 14:44
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