How many different 3-digit even numbers can be formed so that all the

Question

How many different 3-digit positive even numbers can be formed so that all the digit are different?

A. 72
B. 128
C. 256
D. 328
E. 450

ashudhall · Answer

Let's take the 3 ,digit even number which does not have 0 as last digit
Possible numbers = 8*8*4 = 256
3 digit even number with last digit 0 = 9*8 = 72 
Adding both
328
Hence D

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Archit3110 · Answer

total even digits ; with 0 at units place ; 9*8*1 ; 72
total even digits with 0 at tens place; 8*1*4 ; 32
and total even digits without 0 ; 8*7*4; 224
total ; 224+32+72 ; 328 
IMO D

avin2788 · Answer

Can you tell me where I am wrong -

Since, hundreds digit selection can be done from digits 1 to 9,

Hundreds digit selection - 9C1 = 9

Tens digit from 0 to 9 with not repeating the /hundreds digit -

Tens digit selection - 9C1 = 9

Unit digit selection from 0,2,4,6,8 - 5 different outcomes.

Hence - 9*9*5 = 405.

Bunuel  . pls help.

phankism · Answer

Total number of 3-digit even number include 0 at the hundreds digit = 5*9*8 = 360
Total number of 3-digit even number with 0 at the hundreds digit = 1*4*8 = 32
--> Answer = 360 - 32 = 328 --> D

GMATinsight · Answer

The unit digits may be 0, 2, 4, 6, 8

Case 1  : Unit digit of the three digit even number is 0

Unit place can be filled in 1 way (by digit 0)
Tens place can be filled in 9 ways (by any digit other than 0)
Hundred's place can be fille din 8 ways (By any digit other than used at unit's and ten's place)
Total numbers = 1*9*8 = 72

Case 2  : Unit digit of the three digit even number is one of the {2, 4, 6, 8}

Unit place can be filled in 4 way (by digit 2 or 4 or 6 or 8)
Hundred's place can be filled in 8 ways (by any digit other than 0 and the unit digit chosen)
Ten's place can be fille din 8 ways (By any digit other than used at unit's and Hundred's place)
Total numbers = 4*8*8 = 256

Total Favorable Numbers = 72+256 = 328

Answer: Option D

Karmesh · Answer

The units digit needs to be even, which allows us 5 possible values.
Now, Since 1st digit can not be zero and also can't be the same as the 3rd digit, we have 8 ways to choose it.
Similarly, 2nd digit can not be equal the same as the 1st, 3rd digits, hence we can select it in 8 ways. 
 8   8   5  = 320 
This is 8 short of the answer.

I understand the solutions given above, BUT
I need to understand why this is not right? why do we have to make cases for the position of zero to solve it?

Bunuel   VeritasKarishma  If you could please shed some insight

chetansood · Answer

XY0 X=9 Y=8 Number choices at positions of x and y
XY2 X=8 Y=8
XY4 x=8 y=8
XY6 x=8 y=8
XY8 x=8 y=8

Total 64*4+ 72 =328

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chetansood · Answer

With your methode when u say first cant be the same as last then u assume first will be even as the last one is even also.

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Karmesh · Answer

I am not sure what you mean, but, i'll try to explain from what i understood you meant to say.

When i say 1st cant be the same as the 3rd and hence has 8 options. I don't assume the 1st to be even aswell, if i did i would only have 4 options to choose from namely 2,4,6,8. But, since i am taking 8 possible options for the 1st digit, i include the possibility of odd numbers. 
I do assume that we are not selecting the even number that has already been selected. 
Does it make sense?

KarishmaB · Answer

Look at the highlighted part: What if the third digit is 0? In how many ways can we pick the first digit then? In 9 ways. 
But if the third digit is not 0, then there are 8 ways of choosing the first digit. 
Hence you need to take two different cases into account: 0 in units place, any other of the 4 even digits in units place.

djangobackend · Answer

4 cases exist - E= even, O =odd

EEE - 4*4*3 
OEE - 5*5*4 
EOE - 4*5*4
OOE - 5*4*5 
total = 48+100+80+100= 328

How many different 3-digit even numbers can be formed so that all the

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