Of the three-digit integers greater than 660, how many have two digits

You are missing 677, 688 and 699.

gosh, only I can do such stupid mistakes! all efforts are in vain, if u come to the wrong answer at last.
thnx Bunuel. appreciate ur help. next time I will be more attentive (at least I hope so)

from 661 to 699
66_ ( you have 8 numbers )
6_6 ( you have 3 numbers )
_66 ( you have 3 numbers)
Now from 700 to 799
There are 99 numbers in all. In this 72 numbers (9x8) do not satisfy the conditions
therefore, 99-72 =27 numbers satisfy the conditions
Hence, 27x3+14 = 95 is the answer.

I tried to use Bunuel's method, but it was hard for me to escape minor mistakes, consequently my answer wasn't correct.

I find the following way is easier to remember:

600 700 800 900
abc 3 9 9 9
abc 8 9 9 9
abc 3 9 9 9

9*10=90-1+6=95

Hello Bunuel, Considering this logic , can we say there would 9*8*7 three digit numbers with distinct digits

There are 9*9*8 3-digit numbers with distinct digits.

9 options for the first digit (from 1 to 9 inclusive).
9 options for the second digit (from 0 to 9 inclusive minus the one we used for the first digit ).
8 options for the third digit (from 0 to 9 inclusive minus 2 digits we used for the first and the second digits)

700 , 800 , 900 are missing you have to consider these as well



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Hi Bunuel can you kindly elaborate on the first part of the "all the digits are distinct" I didn't quite understand the 3*8, however, I do understand the second part.

Thanks!

Three-digit numbers greater than 660 with the hundreds digit of 6: 6XY - X can take only 3 values (7, 8, or 9) and Y can take 8 values (Out of 10 digits, not 6 and not the onte we used for X).

Let us first take 700-799

With tens digit 7 we have 9 cases. With units digit 7 we have 9 cases. With tens and units digits same we have 9 cases for a total of 27 cases. Similarly for 800-899 and 900-999 for a total of 81 cases.

Let us now take 661-699

With tens digit 6 , we have 8 cases. With units digit 6, we have 3 cases and with units and tens digit same. we have 3 cases for a total of 14 cases.

So for numbers greater than 660 we have 81+14=95 cases

Bunuel Thanks!

Let's breakdown the problem into 2 sections:
i) 661 - 699:
case I: x x y
- - - x -> takes only one value i.e., 6 and y takes 9 digits (0,1,2,3,4,5,7,8,9. Since we are already using up 6) It becomes 1*1*9=9
case II: x y x
Applying the same rule as above, we again get 9.
Hence 9+9 numbers between 661-699

ii) Consider 700-999

X X Y => X can take 3 numbers (7,8,9) and Y can take 9 numbers, giving 3*1*9 = 27
X Y X and Y X X=> Same logic as above we get 27*3=81

i) + ii) = 18+81 = 99

Now out of 99, we have 666,777,888,999 that can not be considered.

99-6=95

There are three possibilities:
a) We need the units digit to be 1 and above keeping the other 2 digits 66 - Clearly there are 8 options (661, 662, 663, 664, 665, 667, 668 and 669) ... (i)
b) We need the tens digit to be 7 and above keeping the hundreds digit 6 -
One scenario is that the tens and units digits are same: 677, 688, 699 - 3 options ... (ii)
Another is that the hundreds and units digits are same: 676, 686, 696 - 3 options ... (iii)
c) We need the hundreds digit to be 7 and above (3 options - 7, 8 and 9) -
One scenario is that the tens and units digits are same for each value of the hundreds place: 3 * 9 * 1 = 27 (i.e. 700, 711, ... 799, 800, etc.) - we chose 9 since the digits 7 cannot be used in the tens and units place and similarly for 8 and 9) ... (iv)
Another is that the hundreds and tens places are equal (units is different) - 3 * 1 * 9 = 27 ... (v)
Another is that the hundreds and units places are equal (tens is different) - 3 * 9 * 1 = 27 ... (vi)

Thus, total = 8 + 3 + 3 + 27 + 27 + 27 = 95
Answer D

what about 661 to 699?

where we add that?

Your question is not clear. What do you mean by referencing the numbers 661 to 699? These numbers are already included in the total count of numbers (95), which have two digits that are equal to each other and the remaining digit different from the other two. Why should these particular numbers be singled out or treated separately?