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 am trying to figure out the maximum possible combinations of a (HEX) string, with the following rules:

  • All characters in uppercase hex (ABCDEF0123456789)
  • The output string must be exactly 10 characters long
  • The string must contain at least 1 letter
  • The string must contain at least 1 number
  • A number or letter can not be represented more than 2 times

I am thinking the easy way to go here (I am most likely wrong, so feel free to correct me):

  1. Total possible combinations: $16^{10} = 1,099,511,627,776$
  2. Minus all combinations with just numbers: $10^{10} = 10,000,000,000$
  3. Minus all combinations with just letters: $6^{10} = 60,466,176$
  4. etc...

Can someone could tell me if this is the right way to go and if so, how to get the total amount of possible combinations where a letter or a number occur more than twice.

Any input or help would be highly appreciated!

Muchos thanks!

PS.

I don't know if I tagged this question right, sorry :(

DS.

share|improve this question
    
I changed it to combinatorics, which is all about counting things. The thing is, you use combinatorics often in probability, so your choosing of probability wasn't a bad tag. I just thought combinatorics was more accurate. This problem could be done in relation to some probability, or without any relation to probability. –  Graphth Feb 27 '12 at 21:39
    
It's not clear if you are counting different combinations (order does not matter) or numbers of strings (order matter) –  leonbloy May 1 '12 at 1:14
add comment

1 Answer 1

I think the sanest way to split this is by how many different letters and numbers is contained in the string. Call the total number of different symbols $k$. The possibilities are then:

k=5    1+4  2+3  3+2  4+1
k=6    1+5  2+4  3+3  4+2  5+1
k=7    1+6  2+5  3+4  4+3  5+2  6+1
k=8    1+7  2+6  3+5  4+4  5+3  6+2
k=9    1+8  2+7  3+6  4+5  5+4  6+3
k=10   1+9  2+8  3+7  4+6  5+5  6+4

For each of the 33 entries in the table, the number of ways to select which hex digit appear at all is a simple product of binomial coefficients.

From there each row of the table can be summed and treated uniformly. First select which $10-k$ of the $k$ symbols are going to appear twice, for a factor of $\binom{k}{10-k}$. Then multiply by $10!$ for the ways to arrange the symbols you have selected, and divide by $2^{10-k}$ to account for the fact that each pair of two equal symbols cannot be distinguished.

share|improve this answer
    
Thanks Henning, I appreciate your effort to help me! However, this answer is waaaay over my head. I didn't even know what word or symbol to google first :( –  José Feb 27 '12 at 21:52
add comment

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.