Alternative Approach:
The required numbers can be counted as follows:
The 2 digits can be selected out of 0, 1, 2…9 in 10C2ways = 45.
Out of these 2 selected numbers, the number which occurs twice can be selected in 2C1 ways = 2.
Now we need to arrange the digits to form number which has 2 identical digits, 1 different digits. It can be arranged in 3!/(2!*1!)= 3 ways.
Total numbers = 45*2*3 = 270.
But this also includes the cases which has 0 at hundredth place, we need to eliminate these cases:
001, 002, 003, …, 009 = 9
010, 020, 030, …, 090 = 9
011, 022, 033,…, 099 = 9
the cases which has 0 at hundredth place = 9*3=27
Required numbers = 270 - 27 = 243. Answer C
MathRevolution wrote:
[GMAT math practice question]
How many different three-digit numbers can be formed which contain two digits that are the same, and a third digit that is different from the other two?
A. 196
B. 216
C. 243
D. 256
E. 316
=>
These three-digit numbers can have one of the forms XXY, XYX and YXX.
Note that 0 cannot be the hundreds digit.
Case 1): XXY
There are 9 possibilities for X (X is not 0), and 9 possibilities for Y (Y≠X).
This gives a total of 9*9 = 81 possible three-digit numbers of this form.
Case 2): XYX
There are 9 possibilities for X (X is not 0), and 9 possibilities for Y (Y≠X).
This gives a total of 9*9 = 81 possible three-digit numbers of this form.
Case 3): YXX
There are 9 possibilities for Y (Y is not 0), and 9 possibilities for X (X≠Y).
This gives a total of 9*9 = 81 possible three-digit numbers of this form.
Thus, the total number of possible three-digit numbers is 81 + 81 + 81 = 243.
Therefore, C is the answer.
Answer: C
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