## A 3-digit positive integer consists of non zero digits. If

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### A 3-digit positive integer consists of non zero digits. If

by BTGmoderatorLU » Thu May 09, 2019 11:55 am

00:00

A

B

C

D

E

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Source: GMAT Prep

A 3-digit 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

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by ceilidh.erickson » Thu May 16, 2019 12:08 pm
There are 3 configurations that would yield a 3-digit 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 non-zero 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.
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by GMATGuruNY » Thu May 16, 2019 1:55 pm
BTGmoderatorLU wrote:Source: GMAT Prep

A 3-digit 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
Alternate approach:

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 1-9 other than the digit already used.)
Number of options for the units digit = 7. (Any digit 1-9 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(81-1-56) = 9(24) = 216.

The correct answer is C.
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by [email protected] » Thu May 16, 2019 6:52 pm
BTGmoderatorLU wrote:Source: GMAT Prep

A 3-digit 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
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.