Originally Posted by

**Debsta** Actually I am now rethinking the solution for (c).

For the number to be even, it must end in 0, 2,4,6,8.

If it ends in 0, there are 9 ways to fill the first box.

BUT if it ends in 2,4,6,or 8, there are only 8 ways to fill the first box.

So we need to count how many end in 0, separately to those that end in 2,4,6,or 8.

End in 0:

?x?x?x?x?x1 (one possibility (ie zero) for the last box)

__9x8x7x6x5x1__ (then fill in the others from the beginning)

End in 2,4,6 or 8:

?x?x?x?x?x4 (4 possibilities at the end)

8x?x?x?x?x4 (8 possibilities for the first - can't be 0 or the end number)

8x8x?x?x?x4 (8 possibilities left for the second number)

__8x8x7x6x5x4__ (fill in the others)

So adding these (underlined) values will give the number of numbers you want (ie the 6-digit numbers that end in 0 and the numbers that end in 2,4,6,or 8, with no repeated digits.)

Sorry about the bum steer with (c) before.