How many even 3 digit numbers can be formed from digits 0 1 2 4 5 6 7 8 9 if each digit can be used only once?

Answer 1

I've thought about this, and with your kind permission, I need to put 'zero' aside for later.

That leaves us eight digits to work with . . . 1, 2, 4, 5, 6, 7, 8, and 9.

Now to build 3-digit even numbers:

-- The last digit can be any one of four. (2, 4, 6, or 8) For each of those . . .

-- The first digit can be any one of the remaining seven. For each of those . . .

-- The middle digit can be any one of the remaining six.

Total number of possibilities so far = (4 x 7 x 6) = 168 .

Now to deal with that 'zero' that I had put aside . . .

For each of the 168 possibilities so far . . .

-- I can leave the 3-digit number exactly as it is.

-- I can substitute the 'zero' for the last digit, and have a different even number.

-- I can substitute the 'zero' for the middle digit, and have a different even number.

-- But I can't substitute the 'zero' for the first digit.

So, for each of the original 168 possibilities, there are three different ways to

introduce 'zero' into the picture, producing three different 3-digit even numbers.

So the grand total of all possibilities is

3 x 168 = 504

And that's the way I see it.

Thank you.