How many even four-digit numbers can be formed from the digits 0,1,2,5,6, and 9 (repetition not allowed)?

An ordering where position matters is a permutation. With repetition disallowed:

In general, nPr(n,r) = n!/(n-r)!

So 6!/(6-4)! = 6!/2! = 360

Or simply nPr(6,4) = 360

First let's see how many numbers you get where the 0 is at the end.

Then the third digit could be 1 out of 5, the second digit 1 out of 4 and the first digit 1 out of 3

So 5*4*3 = 60

Then look at all numbers where the 0 is at position 3

The 4. digit could be 1 out of 2 (2 or 6 since it has to be even), 2. digit is 1 of 4 and first digit 1 of 3

2*4*3=24

When the 0 is at position 2 you get the same result 2*4*3=24

The 0 can never be at position 1 so the final result is 60 + 24 + 24 = 108

edit:

I forgot to add the numbers that do not have a 0 in them at all.

That's 2 * 4 * 3 * 2 = 48

So the total result is 156

But I guess there is an easier way to get the result

First you calculate how many possible combinations there are where the last digit is even. 3*5*4*3=180

And then you subtract the ones that start with a 0. 1/3 of all numbers end with 0 so they are out already.

Of the 2/3 that don't end in 0, 1/5 has to start with a 0 since there are are 5 possible digits they could start with. So 1/5 of 2/3 of 180 is 24 and 180 - 24 is 156.

720

I counted a total of 6 different numbers and pressed the "x!" button on my calculator

(6x5x4x3x2x1)