# How many four digit numbers are there?

Assume that 0 can't be a first digit. I got 9,000. Is that right?

Follow up question: How many of those four digit numbers have no repeated digits?

• A way to see why the first one is correct (posting as comment since what I have isn't a full answer); note that 1000 is the smallest 4-digit number, 9999 is the largest 4-digit number, and all numbers in between are 4-digit numbers. Dec 5, 2013 at 18:34

Yes, you are correct on the first question.

For the second, there are four positions, the first of which has $9$ possible values, (can't be zero), then the second position has 9 possible values it can take on (subtracting 1 from 10 possible values since it can't be the same number as the first position. Etc...

$$9 \cdot 9\cdot 8 \cdot 7 = 4536$$

• ++++++++++++1 :-) Dec 6, 2013 at 2:24

$$9000$$ is correct since the highest number (4-digit) is $$9999$$, and the lowest is $$1000$$. $$9999-1000+1=9000$$. The plus one comes from adding back the excluded item. Let's say there are two numbers, $$1$$ and $$2$$. $$2-1+1=2$$ numbers.

Equivalent question: You have a bag with 10 marbles in it labeled 0 through 9. You pick out a marble 4 times and place it back in the bag each time. The first time, there is no marble with a 0. The total combinations of picking and replacing 4 marbles with the no-zero-condition on the first pick is equal to the product of how many marbles are in the bag for each pick:

$$9 \times 10 \times 10 \times 10 = 9000.$$

For no repeated digits, it's the same problem, but you don't replace the marble (but on the second pick, you place the 0 marble back in):

$$9 \times 9 \times 8 \times 7 = 4536$$

In order to count four digit numbers not beginning with $0$ with no repeating digits, proceed one digit at a time.

There are $9$ possibilities for the first digit. Since the second digit must be distinct, there are $10 - 1 = 9$ possibilities. For the third digit, we have $10 - 2 = 8$ possibilities, and for the fourth, we have $7$.

This yields $$9 \cdot 9 \cdot 8 \cdot 7 = 4536$$ possible numbers.

Your first answer is true, 9000. Since you can only choose 9 digits of the first one, while the second, third, and fourth have 10 choices, that is, $9*10*10*10=9000.$ However, for the second problem, since the first digit have 9 choices, but the second one have 9, the third have 8, the fourth have 7, so the sum is $9*9*8*7 = 4536$.

Yes, $$9000$$ is correct.

For your second question, there are $$4536$$ numbers. $$9\times9\times8\times7$$