# Compute how many n-digit numbers

1. Sep 3, 2013

### knowLittle

1. The problem statement, all variables and given/known data
Compute how many n-digit numbers can be made from the digits of at least one of {0,1,2,3,4,5,6,7,8,9 }
Assume, repetition or order do not matter.

2. Relevant equations
$a_{1}, a_{2}, ..., a_{n}$

3. The attempt at a solution
10 choices for the 1st sub-index, 10 choices for the second sub-index, ..., 10 choices for the nth- sub-index.
$10^{n}$ total possible combinations.
I think that now we need to add 'n' for a set full of one identical digit. i.e.: {2,2,...,n-th}
Now, n*9 for all possibilities

Now, I need to take some n_i number into account in pairs and each of the n-2 numbers repeated for each number.
So, groups of two repeated for each i-th number?
Also, then I would extend this to include triple identical numbers and the rest (n-3) numbers in the set?
And, so on...?

I am sorry, if this does not make any sense or it is too messy.
Could someone give me any guidance?
Thank you.

2. Sep 3, 2013

### LCKurtz

How can order not matter for numbers?

Is it OK to have a string of 0's for leading digits?

I don't follow what you are doing in this last section. Is there something about the problem you haven't told us?

3. Sep 4, 2013

### knowLittle

The question is for a Theoretical Computer Science class.
You might be right, but he said that it does not matter. My professor was very vague in posing the question. I asked it twice.
He gave some examples:
${0, 0, 0,0, ...,0_{n} }={0}$
${0, 0, 0, ..., 0, 1_{n}}={0,1}$

Yes, it is.
This counts as a valid number.
${ a_{0}, a_{1}, ..., a_{n} } \equiv {0_{0},0_{1},...,0_{n} }$

Sadly, I have told you everything that was given to me.

In the last section, I wanted to start counting numbers as if repetition mattered, but I recalled that the professor said that they do not.

I will for sure ask for more clarity next time, I see him.

4. Sep 4, 2013

### LCKurtz

OK. Since leading zeroes are OK it seems to me that your original calculation of $10^n$ should do it.

5. Sep 4, 2013

### knowLittle

Let me get back to you in a few days, I believe I can obtain more specific information about the problem. Thank you for your help so far.

6. Sep 10, 2013

### knowLittle

It turns out that order does matter after all. One 'gotta' love language.
An example helped to elucidate.

Example:
{0,1,2,3,...,8,9, ...<whatever>,...} is a vector of size n and 1 solution that satisfies the constraints.
{3,4,7,...,9,2,1,...<whatever>,... } is another vector of size n and another solution.

Let me work it out. If you have any leads, they are welcome.
Thanks :>

7. Sep 16, 2013

### knowLittle

Please, check the solution in attachment.
Apparently, it is incorrect. Can someone verify?
I think that I am not taking into account cases such as {m, o ,m } or {1,0,1}, where there could be repetitions.

The solution should be in the form:
Order matters
{ ... , <at least digits from 0 to 9>,..., <any numbers>, ... }

Thank you.

#### Attached Files:

• ###### theorCompScie.gif
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8. Sep 22, 2013

### knowLittle

So, here is the solution.

$\dfrac{ |n| !}{ |n_{1}|! |n_{2}|!... |n_{k}|!} \text{, where n is the size of the vector and the values in denominator are types of symbols in n that repeat.}$ $\text{For instance, if I have a vector called v={e, i, g, e, n, v, a, l, u, e}, there are say n1 type symbols.}$ $\text{n1 relates to, say, e and our |n1|=3. The numerator in the equation takes care of all combination including repetitions}$ $\text{and the denominator takes care of cases as the one just mentioned. }$