math
posted by greeny on .
how many different positive fourdigit integers can be formed if the first digit must be 2, the last digit cannot be zero, and digits may be repeated?
I know the answer is 900 but I'm not sure how I would get the answer? thanks!

If the first digit must be 2, and digits can be repeated, you can forget about the leading digit and just look for every THREEdigit integer (INCLUDING leading zeros) whose last digit isn't zero. That's every integer between 000 and 999 (of which there are 1000), excluding the 10% of them that end in zero (of which there are 100)  which makes 900 in total.

Only 1 choice for the 1st spot, 10 choices for the 2nd, 10 choices for the 3rd and 9 choices for the 4th
1x10x10x9 = 900