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PS - Number Prop

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Senior Manager
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PS - Number Prop [#permalink] New post 04 Jan 2009, 12:35
00:00
A
B
C
D
E

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  5% (low)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
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
CEO
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Re: PS - Number Prop [#permalink] New post 04 Jan 2009, 13:01
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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Re: PS - Number Prop [#permalink] New post 04 Jan 2009, 13:13
is the answer C (80)
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Re: PS - Number Prop [#permalink] New post 04 Jan 2009, 13:44
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, 13:44
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