# Thread: how to determin number of combinations

1. ## how to determin number of combinations

hello...

I'm having a bit of problem with one simple task First sorry if I put this thread in wrong place

I know how to calculate number of combinations (without repetition)

$\displaystyle \frac {n!}{k!(n-k)!}$

and that's ok

but now I have big problem I have to calculate number of combinations with repetitions so i tried like this

when i have 2 variables A and B with second expansion i have :

$\displaystyle AA, AB, BA, BB$

so it's four combinations

for third expansion i have :

$\displaystyle AAA, AAB, ABA, BAA, BBB, BBA, BAB, ABB$

so there is 8 combinations .... and at first i think formula should be $\displaystyle n^k$ or in my case now $\displaystyle 2^n$ meaning $\displaystyle 2^2 = 4 and 2^3 = 8$ but as I get to problem with 4 variables and need to do third expansion by that formula I need 64 combinations... I wrote all of them and I got 40... now I'm confused very much because in next task I have problem with 10 variables and third expansion ... and there is no way that i can wrote them all to find pattern ....

can anyone please help me with this I also find some formulas for similar things but none of them can be any use to me in this problem

2. here is how i got combinations with 4 variables and third expansion

$\displaystyle \begin{matrix} AAA & BBB & CCC & DDD & AAC & BBD\\ AAB & BBC & CCD & DDA & ACA & BDB\\ ABA & BCB & CDC & DAD & CAA & DBB\\ BAA & CBB & DCC & ADD & ACC & BDD\\ ABB & BCC & CDD & DAA & CAC & DBD\\ BAB & CBC & DCD & ADA & CCA & DDB\\ BBA & CCB & DDC & AAD & & \end{matrix}$

so there are 40 combinations

I think you mean if there are ten symbols, ABCDEFGHIJ, how many strings of length ten can be formed using any or all of the ten.

For example: $\displaystyle FJBBIIAGHI\text{ or }JIHGFEDCBA$ are two possible strings.
In that example there are $\displaystyle 10^{10}$ possible strings.

In general, if you have $\displaystyle \mathif{N}$ different symbols the number of possible strings of length $\displaystyle \mathif{K}$, using any or all of the symbols, is $\displaystyle \mathif{N}^{\mathif{K}}$

If you mean something else, please try to make it clearer.

It's more like how much combinations can U wrote with 10 symbols in string with length of 3 with repetitions

5. Originally Posted by sedam7
It's more like how much combinations can U wrote with 10 symbols in string with length of 3 with repetitions
The answer to that question is just $\displaystyle 10^{3}=10\cdot 10\cdot 10$.

In the case of four symbols it is $\displaystyle 4^3=64$.

6. Originally Posted by sedam7
hello...

I'm having a bit of problem with one simple task First sorry if I put this thread in wrong place

I know how to calculate number of combinations (without repetition)

$\displaystyle \frac {n!}{k!(n-k)!}$

and that's ok

but now I have big problem I have to calculate number of combinations with repetitions so i tried like this

when i have 2 variables A and B with second expansion i have :

$\displaystyle AA, AB, BA, BB$

so it's four combinations

for third expansion i have :

$\displaystyle AAA, AAB, ABA, BAA, BBB, BBA, BAB, ABB$

so there is 8 combinations .... and at first i think formula should be $\displaystyle n^k$ or in my case now $\displaystyle 2^n$ meaning $\displaystyle 2^2 = 4 and 2^3 = 8$ but as I get to problem with 4 variables and need to do third expansion by that formula I need 64 combinations... I wrote all of them and I got 40... now I'm confused very much because in next task I have problem with 10 variables and third expansion ... and there is no way that i can wrote them all to find pattern ....

can anyone please help me with this I also find some formulas for similar things but none of them can be any use to me in this problem
If you have 2 things (as letters) and n places to put them (as in 2 or three letter words) then you have 2 choices n times. The number of possible "words" is $\displaystyle 2^n$

More generally, if you have m objects and n places to put them, there are $\displaystyle m^n$ possible combinations.

7. Originally Posted by Plato
The answer to that question is just $\displaystyle 10^{3}=10\cdot 10\cdot 10$.

In the case of four symbols it is $\displaystyle 4^3=64$.
hehehehehe... yes $\displaystyle 4^3 = 64$ but up there in post #2 i wrote all of them and there are just 40 combinations or i made mistake

8. Originally Posted by sedam7
hehehehehe... yes $\displaystyle 4^3 = 64$ but up there in post #2 i wrote all of them and there are just 40 combinations or i made mistake