Question 230672
(NOTE: This solution includes a correction to the one provided earlier. My earlier solution mistakenly included 6 among the possible digits. However the new answer still does not match the proposed solution of 456. If the problem is to find the number of all possible even, 4-digit numbers which do not start with zero and which are formed only using 0, 1, 2, 3, 4, 5 as digits, then I believe my solution is correct.
The only ways 456 can be correct is if I have misunderstood the problem or if there are additional conditions which are not specified.
In the message I received from the poster, there is a statement: "It has to be a two case solution." This is followed by "One is to factor the zero in the last position and then case 2 is to take the remaining choices with other 2 even ex zero." which makes no sense to me. There is no reason I can see, given the problem as stated above, to consider a zero in the last (ones) position as a separate case.)<br>
The conditions:<ul><li>Digits are from the set {0, 1, 2, 3, 4, 5}</li><li>4 digits (without a leading 0)</li><li>The number must be even (i.e. it must end in an even digit)</li></ul>
So there are 5 possible first digits (all but 0).
There are 6 possible numbers for the 2nd and 3rd digits
And there are only 3 possible numbers (0, 2 and 4) for the last digit.
Since the order makes a difference, this is a permutation. We just multiply the number of possible choices for each digit:
5*6*6*3 = 540