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Counting Numbers -1

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by GmatKiss » Sat May 19, 2012 1:04 pm
1) Of the three-digit integers greater than 600, how many have two digits that are equal to each other and the remaining digit different from the other two?

(A) 120
(B) 116
(C) 108
(D) 107
(E) 72
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by neelgandham » Sat May 19, 2012 1:26 pm
Total number of three-digit numbers greater than 600 = 399.

Total number of three-digit numbers greater than 600 and with all distinct digits = 4*9*8 = 288. (First digit can take 4 values: 6, 7, 8 or 9, and the second digit can take 9 values and third digit can take 8 values.

Total number of Numbers greater than 600 and with all digits the same = 4 (666, 777, 888, 999).

Total number of Numbers with two digits equal to each other and the remaining digit different from the other two = Total number of three-digit numbers greater than 600 - Total number of three-digit numbers greater than 600 and with all distinct digits - Total number of Numbers greater than 600 and with all digits the same

Total number of Numbers with two digits equal to each other and the remaining digit different from the other two = 399-288-4=107.

Answer: D.
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by kullayappayenugula » Mon May 21, 2012 8:57 pm
hi Neelgandham,

I have a doubt regarding the foramtion of "three-digit numbers greater than 600 and with all distinct digits"

it is given as 4*9*8 = 288. But this includes numbers like 661,662,663,.....771,772,... etc.

can you please tell me what is wrong in my understanding here.
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by Stuart@KaplanGMAT » Mon May 21, 2012 9:22 pm
kullayappayenugula wrote:hi Neelgandham,

I have a doubt regarding the foramtion of "three-digit numbers greater than 600 and with all distinct digits"

it is given as 4*9*8 = 288. But this includes numbers like 661,662,663,.....771,772,... etc.

can you please tell me what is wrong in my understanding here.
Hi!

Because you're multiplying by a smaller number, you've actually removed those duplicates.

Here's another way you could write the product:

(number of digits that could go in the first spot) * (number of digits that could go in the second spot that you haven't used yet) * (number of digits that could go in the third spot that you haven't used yet)

In the first spot, we have 6, 7, 8 and 9, for a total of 4 possible digits.
In the second spot, you have all 10 digits available (don't forget about 0); however, since you used one of them up in the first spot, there are only 9 digits left.
In the third spot, you have all 10 digits available; however, since you used one of them up in the first spot and a different one up in the second spot, there are only 8 digits left.

Accordingly, the number of non-digit repeating integers you can create is:

4*9*8 = 288
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