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by Anurag@Gurome » Mon Jun 18, 2012 10:59 pm
hey_thr67 wrote:How many different three-digit numbers contain both the digit 2 and the digit 6?
In 3 places, we have to place 2, 6, and a third digit from 0 to 9.
Now, for placing 2 we have 3 positions. After placing 2, we have 2 positions left for 6. And after placing 6, we are left with one position to place any of the 10 digits.

Hence, number of integers = 3*2*10 = 60

Now, within these 60 integers there are 026 and 062, which are not three-digit integers. Also we are double counting the following integers 226, 262, 266, 622, 626, and 662, i.e. 6 integers.

Hence, total number of different three-digit numbers contain both the digit 2 and the digit 6 is (60 - 2 - 6) = 52

The correct answer is A.
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by GMATGuruNY » Tue Jun 19, 2012 2:42 am
How many different three-digit numbers contain both the digit 2 and the digit 6?
(A) 52
(B) 54
(C) 56
(D) 60
(E) 62

OA later
Case 1: 2 digits are the same.
226,262,622,662,626,266 = 6 options.

Case 2: All 3 digits are different.
Number of options for the third digit = 8. (Any digit other than 2 or 6.)
Number of ways to arrange the 3 digits = 3! = 6.
To combine the options above, we multiply:
8*6 = 48.
Bad integers are 026 and 062.
Subtracting the 2 bad integers:
48-2 = 46 options.

Total options = 6+46 = 52.

The correct answer is A.
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