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If I have the digits from $0$ up till $9$: $0,1,2,3,\cdots,9$. How many 3-digit number can be made from these set of digits if the number is greater than $600$?

My solution was as follows:There are $4$ ways to choose the first digit $(6,7,8,9)$, $10$ ways to choose the second digit, and finally $9$ ways to choose the third digit (from $1$ up till $9$) so the total number of $3$-digit numbers formed that are greater than $600$ = $4 \times 10 \times 9 = 360$

But, trying to solve the same problem another way gave me another answer: $601$ is the smallest possible 3-digit number that is bigger than $600$ and the largest possible 3-digit number is $999$ so the total numbers will be $999-601+1 = 399$

I wish I know what's the mistake in the second solution which I suspect to be the culprit in having a contradiction as I trust the first answer. Hope someone explains to me. Thanks!

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    Because the number has to be bigger than$600$so I thought that I should be starting with 601 but I missed the case where$0$is perfectly fine as a third digit if the first digit is 6 & the second digit is 1. Andre & David already explained it well. Thanks for both of them!2012-07-01

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In computing the number of ways to choose the third digit, your first argument is in error. The number of ways to choose the third digit is 9 only if the first digit chosen was 6 and the second digit chosen was 0, it is 10 otherwise (for instance $61x$ has 10 possible selections for $x$). Correcting said error will give you a total of 399 ways to form the number.