Question 547666: how many 4 digit even numbers can be formed if repeats of digits are not allowed
Found 2 solutions by mathie123, andy_mitc:
Answer by mathie123(224)   (Show Source):
You can put this solution on YOUR website! Okay so we have ___ ___ ___ ___ to fill in.
Since the number is even, the last blank must be an even number (0,2,4,6,or 8). So there are 5 choices for the last blank.
For the first blank, we cannot have a 0 (as it would not be a true 4 digit number) and we cannot have the digit that was already placed in the 4th blank. This means we have 8 options.
For the second blank, we can now have 0, but we cannot have either of the two numbers that have already been placed. This means we have 8 options.
Similarly, for the third blank we cannot have any of the other 3 numbers that are chosen so we have 7 options.
Using the product rule we get that there are  options.
Hopefully this helps, let me know if you are still unsure.
Answer by andy_mitc(1) (Show Source):
You can put this solution on YOUR website! Unfortunately the previous answer was slightly wrong.
Consider that when 0 is the final number, the values available would become
9 in the thousands place (1,2,3,4...etc)
8 in the hundreds place (remove the previously used number and 0)
7 in the tens place (remove the two previously used numbers and 0)
Therefore, the possibilities would be 504 for 0, higher than the other values you could have placed.
Considering this, we could assume that the previous answer reflects the number of possibilities that were odd. Therefore we could remove the number of non - repeating possibilities with the non-repeating odd possibilities to find the even possibilities.
(9 * 9 * 8 *7) - (8 * 8 * 7 * 5)
= 4536 - 2240
I believe that the answer would therefore be 2296.
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