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?
 A: 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...
A: 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.
A: 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
$$\binom{10}{5}5!-\binom{9}{4}4!$$
A: 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 $7$
$9*9*8*7$
