Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

2 Answers 2

up vote 1 down vote accepted

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.

I wouldn't be surprised if the actual formulation in the book is more precise. Generally speaking, if you ask questions about something in a book you don't understand or think might be wrong, it's a good idea to quote the book verbatim, since the problem may already be in your (re)interpretation of the question.

share|improve this answer
    
Thanks, so it's a question of sematincs. But I still think the book question is confusing. "A bag has 10 balls numbered 0 - 9. From that bag 4 balls are taken, one by one, placing them in row by order of taking, forming a number. How many diferent numbers is it possible to form?". Unless it has to specifically indicate "number of 4 digits", i fail to see the clear indication of either method. –  T23 Feb 14 '12 at 23:11
    
@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

If zero is leftmost, you still get a number. Three-digit numbers are, after all, numbers.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.