# Thread: How many possible arrangements ...

1. ## How many possible arrangements ...

Hello,
I have a question

I have been trying to investigate the formula that gives the number of possible arrangements of $n$ elements in $m$ cells. But I just can't get this formula. I've been thinking a lot and I think it involves factorials but this is just confusing me since two different arrangements could be equivalent given other formulas

For example, if I have three cells ( $m = 3$), I can arrange two elements ( $n = 2$) in three different combinations.

Can anyone help me out on this formula (it's obviously very simple as I cannot find it ...) ? Thanks all

PS : I have to apply the formula to $n = 8$ and $m = 256$ next. Seems like I'm going to have to count decimal digits lol.

2. Hi Bacterius,

Are you working in binary or generally?

$Mp_N=\frac{2^8!}{(2^8-2^3)!}$ for distinguishable cells.

Are you looking for something beyond that?

3. Originally Posted by Archie Meade
Hi Bacterius,

Are you working in binary or generally?

$Mp_N=\frac{2^8!}{(2^8-2^3)!}$ for distinguishable cells.

Are you looking for something beyond that?
well I was trying to find the formula you just gave me (thanks! ) for a table I had to build, but I thought why not finding the general formula (which I can now deduce from your answer).

Thanks again

4. Hi Bacterius,

the "permutations" or arrangements formula will do that

Arrange all N cells.... N!
Arrange N cells from a total of M cells $Mp_N=\frac{M!}{(M-N)!}$

This formula is just a "tweaking" of the actual situation.

For example, arrange 9 distinguishable cells in all orders...

9(8)7)6(5)4(3)2=9!

arrange only 4 of the 9...

9(8)7(6) just happens to be the same as 9! with the 5! missing
and the 5! can be dispensed with by dividing 9! by 5!
5 is the difference between 9 and 4

so, $9(8)7(6)=\frac{9!}{5!}=\frac{9!}{(9-4)!}$

hence the situation of arranging n from m can be formulated from the M and N using

$Mp_N=\frac{M!}{(M-N)!}$