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!
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.