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?
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1$\begingroup$ 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. $\endgroup$– Dennis MengDec 5, 2013 at 18:34
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6 Answers
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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...
$$9 \cdot 9\cdot 8 \cdot 7 = 4536$$
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$9000$ 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.
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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):
$$9 \times 9 \times 8 \times 7 = 4536$$
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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.
answered Dec 5, 2013 at 18:34
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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 $9*9*8*7 = 4536$.
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Yes, $9000$ is correct.
For your second question, there are $4536$ numbers. $9\times9\times8\times7$
