How many three-digit positive integers with different digits are even?

Question

A. 100
B. 320
C. 324
D. 328
E. 405

Consider all positive integers with three different digits. (Note that zero cannot be the first digit.) Find the number of them which are: (a) greater than 700; (b) odd; (c) divisible by 5

nick1816 · Answer

Case 1- When the unit digit is 2,4,6 or 8.

Total possible numbers = 8*8*4=256

Case 2- When unit digit is 0

Total possible numbers = 9*8*1= 72

Number of three-digit positive integers with different digits are even = 256+72=328

TestPrepUnlimited · Answer

We may use a slot method for this question, create three blanks and we start with the last slot as the units digit must be 0, 2, 4, 6, or 8.

__ * __ * 5

Since we used up one digit for the unit digit, we have only 9 to choose from for the tenths place:

__ * 9 * 5

Similarly, the hundreds place has only 8 options left:

8 * 9 * 5 = 72*5 = 360.

However, we included two-digit numbers among those 360 numbers. We would like to remove all digits that start with 0. Again using the slot method, the hundreds digit must be 0, the units digit must be one of 2, 4, 6, or 8, and then the tenths digit would have 8 options. Thus remove 1 * 8 * 4 = 32 options.

360 - 24 = 328

Ans: D

TarunKumar1234 · Answer

We have to get even 3 digit number, having all digits different.

Case1: when we have zero at last place.
_,_,0 then 9C1*8C1 = 72 ways

Case2: when we have 2,4,6 or 8 at last place.
_,_,2 or 4 or 6 or 8 then 8C1*8C1*4C1 = 256 ways

So, Total = 72+ 256 = 328. So, I think D.

abhinavsodha800 · Answer

Hi Nick,
How can it be 8 * 8 ? r u considering the repated digits ? question says three digit number with different digits . i didn't get why it 8 * 8 *4

abhinavsodha800 · Answer

Since you wrote 8 * 9 * 5 it means you must have assumed that zero cannot be at the first place then why are you saying that in above combination we included two digits number also ?

TarunKumar1234 · Answer

Let me try to explain.

for last place, we have 2,4 ,6 and 8. So, we can select in 4C1 ways. Now, you left with 9 digits.
for 1st place, we can't have Zero, so, we have 8 digits to use in 8C1 ways. Now you left with 8 digits
for 2nd place, we select anyone out of 8 digits in 8C1 ways.

So, we have 8C1*8C1*4C1.

abhinavsodha800 · Answer

Hi Tarun,
i have one more doubt. why can't we consider 0,2,4,6,8 in one go instead of considering two cases ? 
if I choose out of 0,2,4,6,8 then i will have 5 ways for unit digit and 8 ways for tens and 8 ways for hundered digit then my answer is coming as 320 whats wrong in this approach

unraveled · Answer

Abhinav..
You can't do that because you are considering than 0 to be a likely candidate for hundreds place which is then wrong(what's why B is the most answered choice after D).
If you do so then consider this:
Take 0 as units place then all possible cases for hundreds place are 1, 2, 3... 9 except 0, hence 9 possibilities. 
Take any of 2, 4, 6, 8 at units place then all possible cases for hundreds place are 1, 3, 4, 5, 6, 7, 8, 9(when 2 is at units place) and so on, hence 8 possibilities.

Do note that whenever such digits and number problems are there, 0 has to be given special attention - hundreds place if 3-digit number and thousands place if 4-digit number.

rsrighosh · Answer

Bunuel   VeritasKarishma   chetan2u

I did understand Case 1 where unit digit is 0:- that's why I am having 9*8*1 = 72

But I am getting 2 answers for Case 2 which is confusing me. Can anybody help.

Case 2.1 
Unit Digit:- 2,4,6,8 => 4 (Let's say we choose 8)
 Tens  Digit:- 0,1,2,3,4,5,6,7,9 => 9 (Let's say we choose 7)
 Hundreds Digit :- 1,2,3,4,5,6,9 => 7

Hence 4*9*7 = 252

Case 2.2
Unit Digit:- 2,4,6,8 => 4 (Let's say we choose 8)
 Hundreds Digit :- 1,2,3,4,5,6,7,9 => 8 (Lets say we choose 7)
 Tens  Digit:- 0,1,2,3,4,5,6,9 => 8

Hence 8*8*4 = 256

So i I simply change the  sequence  of choosing, I am getting two different answers.

I know obviously I am making a mistake which I am unable to understand. Can anyone please help me out?

chetan2u · Answer

Of course, the correct way is to fill up the places of restrictions first, so case 2.2 is correct => 8*8*4

What is going wrong in 2.1?
What happens when the tens digit is 0, the hundreds digit can be any 8(not just 7)
So within case 2.1, we have 
4*8*7....when tens digit is anything other than 0
4*1*8....when tens digit is 0.

Total
4*8*7+4*1*8=4*8*(1+7)=4*8*8

rsrighosh · Answer

chetan2u I never could have realised there are a subcases in 2.2

Thanks for the tip. Always fill up restrictions first.

HarshavardhanR · Answer

We may have to consider the cases of numbers ending with 0 separately, because

- If a number ends with 0, then there will be 9 different options for the hundredth digit.
- If a number ends with some other even number, then there will be only 8 different options for the hundredth digit (because we cannot have the 3-digit number with 0 as the left-most digit.).

Hence,

Case 1:   Number ending with 0

9 x 8 x 1 = 72.

Case 2:   Number ending with any other even number

8 x 8 x 4 = 256

Final answer: 72 + 256 = 328.

---
Harsha

How many three-digit positive integers with different digits are even?

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