How many 3 digit positive integers with distinct digits

If the second digit is 0, then you'll have 8 not 7 options for the third digit.

Hope it's clear.

I did in this way :

case 1: tens place is 0 : 8 x 1 x 9 = 72

case 2 : tens place is not 0 : 7 x 8 x9 = 504

therefore , total numbers = 72 + 504 = 576 ans.

Bunuel, Why must one go by 1st-3rd-2nd digit option and not 1st-2nd-3rd digit option, which would result in 9*9*8?

hi, there can be certain cases in which zero can occur at the 3rd digit, if one follows 1st-2nd-3rd digit option. in order to eliminate all such cases we go with 1st-3rd-2nd digit option.

remember a number will be divisible by 10 if its last digit ends with zero, and in this case we want to eliminate all such cases in which zero occurs at the unit digit.

i hope it helps.

Hi Manpreet,
The approach is still not clear to me. By 9*9*8 option, we eliminate the possibility of a Zero at the 3rd place.

lets try to understand, why we approach this question from first digit and why not from second or third digit ??

we don't want zero at the first digit. why ?? if zero occurs at the first digit then number won't be a three digit number(it will become a two digit number). that's why we want to eliminate all such cases in which zero occurs at the first digit so we started with first digit. now up to this point we have 9 possibilities for first digit . which are 1,2,3,4,5,6,7,8,9. suppose we selected 9 for the first digit. so number will be 9,_,_

now lets suppose we follow your method. so for second digit we have 0,1,2,3,4,5,6,7,8. now from these 9 numbers suppose we selected digit 8

so the number is 98_

now for third digit we have 0,1,2,3,4,5,6,7, now from these 8 number we can select any of the number. that is we have certain cases in which 0 can also occur at the third digit. (i.e. 980 is one of the possible number) which we don't want.

hence we have to eliminate all such possible cases.

You mixed possibilities with the number at the highlighted section. Let me explain my approach.
1. I have 9 possibilities for the 1st digit.
2. I have again 9 possibilities for the second place as I will consider 0 here
3. Now out of 10, 2 numbers are used. If I assume 0 was used in the second place (10_) then I have 8 possibilities for the last place as well. so the final possibility will be 9*9*8.
I hope I made myself clear on the understanding. Please point out the flaw.

how do you know that zero will be at the second position. ?? i have highlighted one of the cases in which zero will not be at the second position.

and my dear friend, i'm not mixing any possibilities. that's the correct way of approaching the problem.

I found the total distinct#s and subtract the #s that end with Zero
9*9*8 - 9*9*1 = 576

9*9*8 - 9*9*1 = 567 and not 576...

9*9*8-9*8*1=576

_ _ _; Last digit has to be 0, so one possibility.
First digit can be one of 9 possibilities(1 to 9). second digit can be 8 (10-2)

Let’s analyze the number of choices for each of the 3 digits from left to right.

Since the first digit can’t be 0, there are 9 choices for the first digit.

Since the second digit can’t be same as the first digit (but it can be 0), there are also 9 choices for the second digit. However, let’s consider the following two cases: 1) the second digit is not 0, and 2) the second digit is 0. For each of these two cases, we will also consider the last digit.

Case 1: The second digit is not 0

Since the second digit is not 0 (and it can’t be the same as the first digit), there are 8 choices for the second digit. As for the last digit, it can’t be 0 (otherwise it will be a multiple of 10) and it can’t be the same as either of the first two digits; thus, there are 7 choices. Thus, there are 9 x 8 x 7 = 504 three-digit numbers with distinct digits when the second digit is not 0.

Case 2: The second digit is 0

Since the second digit is 0, there is only 1 choice for the second digit. As for the last digit, it can’t be 0 (otherwise it will be a multiple of 10) and it can’t be the same as either of the first two digits. However, since the last digit won’t be 0, it won’t be the same as the second digit. In other words, it only needs to be different from the first digit; thus there are 8 choices. Thus, there are 9 x 1 x 8 = 72 three-digit numbers with distinct digits when the second digit is 0.

Lastly, the number of three-digit numbers with distinct digits is 504 + 72 = 576.

Answer: A

After getting 9 options for the first digit why did you directly jump to third digit and got 8 options which seems correct as we are omitting 0 and the first digit. However after getting the options for first digit if you had moved to selecting options for second digit you would've got 9 options(0-9 minus first digit) and then 7 options for third digit. Please help me understand sir Bunuel

If we directly move to the second digit, and take it as 9 options - we include 0 in this option, and we need to know whether the second digit will be 0 or not, because if it is 0 then the third digit has 8 options and if it isn't 0 then the third digit has 7 options (because anyway third digit can't be 0).

That's why we first jump the third digit and give it 8 options (which ofc doesn't have a 0), and basis that we have 8 options remaining for second digit. Hope it helps.