Source: GMAT Prep
A 3digit positive integer consists of non zero digits. If each exactly two of the digits are the same, how many such integers are possible?
A. 72
B. 144
C. 216
D. 283
E. 300
The OA is C
A 3digit positive integer consists of non zero digits. If
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There are 3 configurations that would yield a 3digit integer with 2 of the same digit, one different digit:
[same][same][different]
[same][different][same]
[different][same][same]
So, let's calculate the number of combinations for one of these:
[same][same][different]
There would be 9 nonzero digits as options for the hundreds digit, and then only 1 option for the tens digit, since it has to be the same as the hundreds digit. You might think that there would be 9 options for the units digit, but remember  it has to be different! So there would be only 8 options that are different from whatever we picked for the hundreds & tens digits.
9*1*8 = 72 options
It would stand to reason that all 3 configurations mentioned above would have the same number of possibilities: 72. So we add:
72 + 72 + 72 = 216
The answer is C.
[same][same][different]
[same][different][same]
[different][same][same]
So, let's calculate the number of combinations for one of these:
[same][same][different]
There would be 9 nonzero digits as options for the hundreds digit, and then only 1 option for the tens digit, since it has to be the same as the hundreds digit. You might think that there would be 9 options for the units digit, but remember  it has to be different! So there would be only 8 options that are different from whatever we picked for the hundreds & tens digits.
9*1*8 = 72 options
It would stand to reason that all 3 configurations mentioned above would have the same number of possibilities: 72. So we add:
72 + 72 + 72 = 216
The answer is C.
Ceilidh Erickson
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Alternate approach:BTGmoderatorLU wrote:Source: GMAT Prep
A 3digit positive integer consists of non zero digits. If each exactly two of the digits are the same, how many such integers are possible?
A. 72
B. 144
C. 216
D. 283
E. 300
Integers with exactly 2 digits the same = Total integers  Integers with all 3 digits the same  Integers with all 3 digits different.
Total integers:
Number of options for the hundreds digit = 9. (Any digit but 0.)
Number of options for the tens digit = 9. (Any digit but 0.)
Number of options for the units digit = 9. (Any digit but 0.)
To combined these options, we multiply:
9*9*9.
Integers with all 3 digits the same:
111, 222, 333, 444, 555, 666, 777, 888, 999.
Number of options = 9.
Integers with all 3 digits different:
Number of options for the hundreds digit = 9. (Any digit but 0.)
Number of options for the tens digit = 8. (Any digit 19 other than the digit already used.)
Number of options for the units digit = 7. (Any digit 19 other than the two digits already used.)
To combine these options, we multiply:
9*8*7.
Thus:
Integers with exactly 2 digits the same = (9*9*9)  9  (9*8*7) = 9(81156) = 9(24) = 216.
The correct answer is C.
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The 3 digits are nonzero, and they consist of two distinct digits. The number of ways to choose 2 digits from 9 nonzero digits is 9C2 = (9 x 8)/2 = 36. Now for each pair of digits chosen, for example, 1 and 2, we could have: 112, 121, 211, 221, 212, and 122. Therefore, there are 36 x 6 = 216 such integers.BTGmoderatorLU wrote:Source: GMAT Prep
A 3digit positive integer consists of non zero digits. If each exactly two of the digits are the same, how many such integers are possible?
A. 72
B. 144
C. 216
D. 283
E. 300
The OA is C
Answer: C
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