how many 5 digit numbers are there with distinct digits?

I found this question on gre forum, it's answer was given by this expression:

$9\cdot9\cdot8\cdot7\cdot6$ which I heard in school as well.

What I tried to do was:

for numbers from index $4$ to index $1$, we can use any of the $10$ numbers $(0-9)$ once so I got this result, $10C4\cdot4$! Now for the first index, it can be not $0$ and $4$ less number or $3$ less number to choose from depending upon whether we are selecting $0$ or not.

I know it is wrong, what Can anyone point out what is wrong with this approach?

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Here the order of digits is important, and what do you need combination or permutation? – Ram Jan 15 '13 at 11:14

This approach is quite hard because of "4 less or 3 less depending upon whether we selected 0 or not" part.

Try to start with the first digit instead. How many choices have you there? How many choices left for the next one? And so on...

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There are 9 choices for the first digit, since 0 can't be used. For the second digit, you can use any of the remaining 9 digits. For the third digit you can use any of the 8 digits not already used. For the next digit, there are 7 choices. And for the final digit there are 6 choices left. Multiplying the values together gives the stated answer: 9 x 9 x 8 x 7 x 6.

Your approach will work, but you need to count how many cases include a 0 in the final four digits, because only these will leave 6 digits available for the first digit. If the last 4 don't include 0, you only have 5 choices left for the first one. Since the number of distinct-4-digits arrangements which don't include a 0 is $^9C_4 \times 4!$, the calculation gives

$\left( ^9C_4 \times 4! \right) \times 5 + \left( \left(^{10}C_4-^9C_4\right) \times 4! \right) \times 6$

which simplifies to $9 \times 8 \times 7 \times (30 + 60 - 36)$, i.e. 9 x 9 x 8 x 7 x 6, as before.

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We choose 5 numbers out of 10 in $\binom{10}{5}$ ways after rearranging in $5!$ ways we get total number of arrangments $\binom{10}{5}5!$ but if 0 is in first place that arrangment is not a five digit number from 9 digts we choose 4 in $\binom{9}{4}$ after arranging them we get $\binom{9}{4}4!$ a 5 digit arrangment that starts with 0 so desired results is $$\binom{10}{5}5!-\binom{9}{4}4!$$

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Write $(0,1,2,3,4,5,6,7,8,9)$ Then the nb of digits (_ , _ , _ , _ , _) First digit there is $9$ choices without $0$ Second is also $9$ choices because we added zero and its permutation Then $8$ and $7$ $9*9*8*7$

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