# Combinatorics and Arrangements

I have a exercise like this in my book:

10 balls numbered 0 - 9, removing 4 balls, one by one, I'll have a number. How many different numbers can I get.

My book has the result as: 10 x 9 x 8 x 7 = 5040
But shouldn't it be 9 x 9 x 8 x 7 = 4536.
I assumed 0 couldn't be the counted when leftmost.

Is the book solution right? I'm confused.

Thanks.

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It depends on the process implied by "I'll have a number". What you'll have will at first be just $4$ balls, drawn in a particular order. You can then form a representation of a number using them, and then you can consider the number represented by that representation. You could concatenate the digits on the balls and then discard the result if it begins with $0$ because that's not a well-formed representation of a number. Or you could associate a string of digits beginning with $0$ with a number in the only natural way, by taking the $0$ to signify $0$ thousands, even though one wouldn't usually write it that way. In the first case, you could get $9\cdot9\cdot8\cdot7$ numbers; in the second case you could get $10\cdot9\cdot8\cdot7$ numbers. This is not a question of combinatorics; it's just that "I'll have a number" is not a very precise description.
@TMorais: You're right, that's not formulated well at all. First, not the balls form a number, but the digits on them. Second, they don't form a number, but a decimal representation of a number. And third, strings with leading $0$s are usually not considered well-formed decimal representations of numbers. So there's nothing to worry about, your understanding is correct and you were just expecting too much precision from the book. Books are only human, you know :-) – joriki Feb 14 '12 at 23:15