rxs0005 wrote:
The Way i solved it is
A B C D ( thousand , hundreds, tens, units)
D can be 0 2 4 6 8 ( any of the 5 digits )
A can be anything except (D or 0) so 8 possibilities
C can be anything execpt A and B so 8 possibilities
B can be anything execpt ( A D C ) so 7 possibilities
total ways are 8 * 7 * 8 * 5 = 2240
The only thing different from this logic and bunuel's way is that this has to be a number that means it cannot start with a 0 as the Q is asking 4 digit even "number"
Since your solution does not include the numbers that start with 0 (0BCD), you should be getting the same answer as Bunuel and Shrouded.
I have spend about 15 mins on this, was trying to figure out
(1) what were you doing wrong
(2) why can't we simply start from the thousandth digit and move to the right.
I believe the way Bunuel wrote is a little simpler & quicker, as long as you understand that you need to calculate all 4 digit numbers and then subtract the one's that start with zero (or an alternative that I will provide below).
(1) what I think is wrong in your approach (you are getting slightly fewer ways ) -
A B C D ( thousand , hundreds, tens, units)
D can be 0 2 4 6 8 ( any of the 5 digits )
A can be anything except (D or 0) so 8 possibilities -
while doing this you have removed extra possibilities. What happens when D itself is 0 ? so you see, some of the A=0 options have already been removed when you said that A can not be same as D. C can be anything except A and B so 8 possibilities
B can be anything execpt ( A D C ) so 7 possibilities
To compensate for the above, you can do this (this is a correction to your approach or can be an alternative to Bunuel's method) -
D = 4 ways (2,4,6,8 only, 0 has been removed)
A = 8 ways ( can't be equal to D and 0)
B = 8 (can't be equal to A & D)
C = 7 (can't be equal to A, B & D)
Gives = 1792 (4 digits numbers without repetition, but without the ones that end with 0)
D = 1 (can only be equal to 0)
A = 9 (anything but D, which also covers cases where A=0)
B = 8 (anything but A &D)
C = 7 (anything but A, B & D)
Gives = 504 (4 digits numbers without repetition that end with 0)
Total = 1792 + 504 = 2296
Brunel - (2) why can't we simply start from the thousandth digit and move to the right - is this because we need to take the possibilities for the restrictive digits first, such as the ones digit that can only be even ?
Edit - realized after posting that Bunuel had posted a reply.