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Counting Numbers -4

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by GmatKiss » Sat May 19, 2012 1:07 pm
Of the three-digit integers greater then 750, how many have at least two digits that are equal to each other ?

(A) 56
(B) 70
(C) 72
(D) 74
(E) 78
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by coolhabhi » Sat May 19, 2012 1:56 pm
IMO : D

Because from 750 the numbers that have at least two digits that are equal to each other are :
755,757 -> 2 Numbers
766,767 -> 2 Numbers
770 to 779 -> 10 Numbers
787,788 -> 2 Numbers
797,799 -> 2 Numbers
---------------------
.........-> 18 Numbers

Now from 800 to 999 there will be 2 numbers for every set of 10 three digit numbers and an additional 10 numbers as seen above.

so it would be 2 * 18 (counted in 800 to 999) + 20 (the 10 numbers for 800 to 999)
=>56
56 + 18 = 74 numbers.
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