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?
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.Sign up to join this community
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...
$$9 \cdot 9\cdot 8 \cdot 7 = 4536$$
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 $9*9*8*7 = 4536$.