1

Please forgive me for this amateur question in advance.

I have 3 letters, A, B & C. I want to find out how many combinations can I make from 1 to 10 characters using these 3 letters.

Say: AAAAAAAAAA, AAAAAAAAAB, AAAAAAAAAC etc.

Where AAAAAAAAAC = AAAACAAAAA = ACAAAAAAAA. (This set is equally the same because of the occurence of a single C)

or say AAAB = ABAA = BAAA.

or say BBBC = BCBB = CBBB.

from 1 to 10 characters.

I hope this is clear. Please help.

N. F. Taussig
  • 76,571
Abu Haydar al-Janoobi
  • 11
  • 2
  • The title mentions "not repeating a set" but this isn't mentioned in the body. What do you mean by it? – redroid Sep 20 '20 at 16:58
  • 1
    I changed the title, sorry I got confused. This thing is giving me a huge headache. Not a mathematician. – Abu Haydar al-Janoobi Sep 20 '20 at 17:01
  • Also to clarify - are you asking how many ways you can make choose 10 letters from the three, disregarding order? – redroid Sep 20 '20 at 17:01
  • from 1 to 10 letters. Like :

    AA AAA AAAA AAABB ... AAAAAAAAAA

     – Abu Haydar al-Janoobi Sep 20 '20 at 17:03
  • Sorry, from 1 to 10 letters, but the important part is that the order doesn't matter. All four of AAAB, AABA ABAA and BAAA should count once, right? – redroid Sep 20 '20 at 17:04
  • An equivalent problem would be, "How many different solutions exist to the equation A+B+C=10 (A,B,C all integers >= 0)". This is turn can be thought of as the sum of 11 simpler questions: "How many different solutions are there to A+B=x", with 0<=x<=10. And that last one is pretty easy. – pokep Sep 20 '20 at 17:04
  • We want sets of the form $A^aB^bC^c$ where $a,b,c$ are the multiplicities of each letter. We need $a+b+c=10$ and $a,b,c$ are non-negative integers ... So ... (Ask if you need more help) $\ddot \smile$ – Donald Splutterwit Sep 20 '20 at 17:04
  • 1
    "Sorry, from 1 to 10 letters, but the important part is that the order doesn't matter. All four of AAAB, AABA ABAA and BAAA should count once, right?" - Yes – Abu Haydar al-Janoobi Sep 20 '20 at 17:07

2 Answers2

3

Hint :

Rephrasing, given four symbols $\{A,B,C,0\}$ how many $10-$lettered combinations possible? $0$ symbol for blank spaces.

Let each symbol occur $a_{i}$ times ($i=1,2,3,4$) and $a_{i}\in{0,1,\ldots10} $. Then solution for $$ a_{1} + a_{2} + a_{3} + a_{4} = 10 $$ is given by $^{10+4-1}C_{4-1}=286$ using stars and bars. Subtracting $1$ from it, to not account for the case when all $0$s used, 285 is final answer.

cosmo5
  • 10,629
0

Because order doesn't matter, your strings are defined by how many of each letter they include. This in turn means your question boils down to, how many ways can I choose integers $A,B,C$ such that $A+B+C=n$ for values of n between 1 and 10?

That question for a specific value of n is answered by this page which says that this value is equal to $\binom{n-1}{2} = \frac{(n-1)!}{2(n-3)!}$.

To answer your question, simply add together all the values of that for $1 \geq N \geq 10$. I get $120$ combinations.

redroid
  • 610
  • 3
  • 13
  • The correct answer is $285$ as given by cosmo5. The value for a specific $n$ is $\binom{n+2}{2}$ by stars and bars. You are applying a formula for solutions in positive integers, but the problem calls for solutions in nonnegative integers. – saulspatz Sep 20 '20 at 17:27
  • So it is, I missed the distinction – redroid Sep 20 '20 at 17:28