how to determin number of combinations

**sedam7** · April 1st 2011, 07:53 AM

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)

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

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 :

$AA, AB, BA, BB$

so it's four combinations

for third expansion i have :

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

so there is 8 combinations .... and at first i think formula should be $n^k$ or in my case now $2^n$ meaning $2^2 = 4 and 2^3 = 8$ 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

**sedam7** · April 1st 2011, 08:37 AM

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

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

so there are 40 combinations

**Plato** · April 1st 2011, 08:40 AM

To be honest I don’t really follow your question.
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: $FJBBIIAGHI\text{ or }JIHGFEDCBA$ are two possible strings.
In that example there are $10^{10}$ possible strings.

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

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

**sedam7** · April 1st 2011, 09:14 AM

sorry for bad explanation...
It's more like how much combinations can U wrote with 10 symbols in string with length of 3

with repetitions

**Plato** · April 1st 2011, 09:22 AM

Originally Posted by sedam7

sorry for bad explanation...
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 $10^{3}=10\cdot 10\cdot 10$ .

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

**HallsofIvy** · April 1st 2011, 09:23 AM

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)

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

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 :

$AA, AB, BA, BB$

so it's four combinations

for third expansion i have :

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

so there is 8 combinations .... and at first i think formula should be $n^k$ or in my case now $2^n$ meaning $2^2 = 4 and 2^3 = 8$ 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 $2^n$

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

**sedam7** · April 1st 2011, 09:27 AM

Originally Posted by Plato

The answer to that question is just $10^{3}=10\cdot 10\cdot 10$ .

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

hehehehehe... yes $4^3 = 64$ but up there in post #2 i wrote all of them and there are just 40 combinations

or i made mistake

**Plato** · April 1st 2011, 09:40 AM

Originally Posted by sedam7

hehehehehe... yes $4^3 = 64$ but up there in post #2 i wrote all of them and there are just 40 combinations

or i made mistake

You made a mistake.

**sedam7** · April 1st 2011, 09:46 AM

Originally Posted by Plato

You made a mistake.

LOL... I'm idiot

i just do it with 2 different variables ....

sorry for all mess that i make .... and thank you all very very much

Thread: how to determin number of combinations

LinkBack

Thread Tools

Display

how to determin number of combinations

Similar Math Help Forum Discussions

Number of combinations.

Number of combinations

determin the symmetry group

number of combinations possible

determin arrangement of points in R3

Search Tags