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

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.

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

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



To attend the !!!FREE!!! Live class on Permutation and combination on 24th July 2019 @9 PM IST, click and register on the LINK

.

.

.

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

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

Posted from my mobile device

[quote="Karmesh"]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.
[u]8[/u] [u]8[/u] [u]5[/u] = 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?

[url=https://gmatclub.com:443/forum/memberlist.php?mode=viewprofile&un=Bunuel][b]Bunuel[/b][/url] [url=https://gmatclub.com:443/forum/memberlist.php?mode=viewprofile&un=VeritasKarishma][b]VeritasKarishma[/b][/url] If you could please shed some insight[/quote]


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.

[size=80][b][i]Posted from my mobile device[/i][/b][/size]

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?

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.

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.