Math question permutations vs exponentiation

[adam2016]

Hi guys,

so really dumb question here but I thought I'd ask it anyway, what is the difference between exponentiation and permutations, so let's say we have 2^8 this will give us 256 bits or 256 possible permutations right?

but the permutation formula is N!/(N-R)! , I subbed 8 in for N and 2 in for R and didn't get 256,

isn't a permutation telling us how many possible arrangements we can have? isn't this not the same as 2^N?

In other words how come 8!/(8-2)! is not == to 2^8?

Probably a really dumb question but nonetheless thanks.

[jonnin]

its the wrong formula :)
the perm formula you are using is for when you take something away, eg you have a sack full coins, and you take 1 coin away to build your perm, then there are n-1 left in the bag, not n.

you need the formula where the magic bag always holds n.
which is.... an exponent :)

the easy one is so trivial that math classes and courses etc tend to gloss over it and favor the second one.

[seeplus]

this will give us 256 bits or 256 possible permutations

256 possible values - not permutations.

A 'simple' definition of permutation is 'designates those ordered arrangements in which no element occurs more than once, but without the requirement of using all the elements from a given set'.

In this context, the important part of the definition is 'without the requirement of using all the elements' and 'ordered arrangements'.

The number of ways of choosing 2 objects from 8 with the ordering of the resultant arrangement important is 8!/(8 - 2)! = 8!/6! = 8 * 7 = 56. So if you had 8 flowers and wanted to choose any 2 were the ordering of the chosen 2 flowers matters, then this can be done in 56 ways.

[jonnin]

right.
you want the answer when you flip a coin 8 times and record the result in order (where heads is 0, tails is 1 for example). Its a very different question -- as I said, the formula you use does not answer the question you posed.

they are both permutations. His simple definition is only one type.

[againtry]

so let's say we have 2^8 this will give us 256 bits or 256 possible permutations right?

The whole problem here is the attempted equivalence is meaningless/incorrect.

2^8 is, in fact, exponentiation - raising 2, in this case to a power, 8 in this case.

And this indeed gives 256 as the answer, but not 256 bits. Pure numbers are scalars, they are dimensionless and a bit is a dimension, or unit if you like.

The result has nothing to do with permutations or combinations.

Now, if you have 8 (binary) bits of information, and this is what this is all about, then all of the possible binary numbers give a binary range from 0000 0000 to 1111 1111 or 0 to 255 decimal. realistically this doesn't have anything to do with permutations or combinations unless you want to waste your time thinking about the permutations of possible values for a single bit - a 0 or a 1 - which goes nowhere.

[seeplus]

re the coin analogy. Consider having 8 coins. Each coin can be either heads or tails. The face of one coin has no bearing on any other coin face.

So the first coin can be arranged in 2 ways - either heads up or tails up. The second can again be arranged in 2 ways - heads up or tails up. The same applies for all of the 8 coins.

Therefore the number of ways that the 8 coins can be arranged differently is:

2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 = 256

ie 2 ^ 8

[againtry]

Correct weight.

A permutation of r numbers from a set S of n numbers with repetition is n^r

- raising n to the power r is exponentiation
- r is the exponent of n

Here, r = 8, n = 2 ( i.e S = {0,1} ) (No single coin zero sum analogy required)
There are 2^8 = 256 permutations, as we all know and have come to love.

[adam2016]

Thanks guys makes sense,

I just used bits as an example to give the question a little meaning I guess, but yeah of course bits is redundant as that part is not important.