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by ketkoag » Tue Mar 31, 2009 10:44 am
Of the three-digit positive integers that have no digits equal
to zero, how many have two digits that are equal to each
other and the remaining digit different from the other two?
A. 24
B. 36
C. 72
D. 144
E. 216

OA: E
I got the right answer that is 9*8*1 + 9*1*8 + 8*9*1
Please lemme know if the above solution is right
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by mike22629 » Tue Mar 31, 2009 11:08 am
yes that approach works for this problem.

9 is possibility of first digit
8 is possibility of second digit
only 1 possiblity for last digit.

Times 3 because there are 3 digits
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by EricKryk » Tue Mar 31, 2009 10:29 pm
So if we knew that the 1st and 3rd digits had to be the same, it would just be 9*8*1? Or would it be 9*8*1 + 1*8*9 ?
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by 2010gmat » Wed Apr 01, 2009 12:06 am
your answer is perfect....

3 digits, can be filled by 2 similar digits in 3 ways, 2 similar digits (barring 0) can be picked in 9 ways...and third digit can be picked in 8 ways

3*9*8 = 216

Let us include 0 and then solve it....
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by vittalgmat » Wed Apr 01, 2009 12:10 am
EricKryk wrote:So if we knew that the 1st and 3rd digits had to be the same, it would just be 9*8*1? Or would it be 9*8*1 + 1*8*9 ?
It would be just 9*8*1.

Let me explain the problem and the solution.
Let us take an example. let us assume that repeated digit is 3 and other digit is 7. So we can make the following combinations.

337, 373, 733 => 3 variations.

The solution is similar to what everyone have solved above.
Lets start from Hundreds position and move to units position.

Hundred's position can be filled in 9 ways coz we have 9 digits (1 - 9).
Ten's position can be filled in 1 way only (coz we want to have the digit in
hundred's position repeated).
units position can be filled in 8 ways coz we dont want the digit that we used for hundred's and ten's positions.

ie. 9*1*8 possibilities for 1 variation.
So for 3 variations we have 9*1*8*3 = 216

Ht helps
-V
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