Counting - How many integers from 100 to 999 . . .

IMO C

Number of digits at 100's place = 9 (0 not allowed)
Number of digits at 10's place = 9 (digit at hundred place excluded and 0 included)
Number of digits at 1's place = 8 (digit at 100's place and 10's place excluded, 0 included)

therefore

9*9*8 = 648

We have a 3 digit number:


- - -

For the first digit, there are 9 choices (1-9 inclusive)

Then for the second digit, there are 9 choices (0-9 inclusive minus the digit used in the hundreds spot).

Then for the 3rd digit, we have 8 choices (0-9 inclusive minus the digit used in the hundreds spot and minus the digit used in the tens spot).

9 * 9 * 8= 648

Answer = C

-Carrie

Brent Hanneson wrote:How many integers from 100 to 999 are such that all three digits are all different?
(A) 504
(B) 567
(C) 648
(D) 720
(E) 729

Hope I'm getting the concept right. Just wondering in a different scenario, if the ques ask for any combination of 2 same digits instead of 3 different digits. Using the same way to solve based on piyush_nitt method, would the result be 9*9*9=729 (E)?

I still have not got this question.

Why to multiply by 8 in the 1st digit?

Thanks in advance!

Maratha1 wrote:I still have not got this question.

Why to multiply by 8 in the 1st digit?

Thanks in advance!

Since the requirement is 3 different digits, therefore the 1st digit has to exclude numbers that is in the tens and hundreds place.

Eg. Let's pick a 3 different digits number - 154.

Number digits available at 100's place are 1,2,3,4,5,6,7,8,9. 0 is not allowed because question number range from 100-999. Therefore, there is 9 digits available.

Number digits available at 10's place are 0,2,3,4,5,6,7,8,9 because digit 1 has been taken by hundreds place so you need to exclude digit 1 which means total number digits available at 10's place is 9.

Number digits available at 1's place are 0,2,3,4,6,7,8,9. Since digit 1 and digit 5 has been taken by the tens and hundreds place that leaves us with 8 digits.

Hope this explanation is helpful.