how many different positive four-digit 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!
2 answers
If the first digit must be 2, and digits can be repeated, you can forget about the leading digit and just look for every THREE-digit 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
1x10x10x9 = 900
1 answer