# How calculate the number of possible different variations?

I feel stupid, but I don't know how to calculate how many possible variations I can get from for example three spaces (1|2|3) Normally I'd say: "well that is easy, just take the number of spaces (=3) and 3^3"

But that doesn't work with two spaces, like "1|2", there are only 2 different ways on how to arrange two numbers, but 2^2 would be 4.

(I want to know how many spaces I need to get ~25000 possible variations)

• What do you mean by "variations" and "spaces"? Maybe you could give some examples to clarify what it is you're looking for? – Hans Lundmark Apr 26 '11 at 9:59
• Don't feel stupid - we all have gaps like this in our knowledge! – Undercover Mathematician Apr 26 '11 at 10:11

By variations, do you mean..

..placement without repetition

There are $n! = n*(n-1)* ... * 3 * 2 *1$ ways of ordering $n$ elements. So for 3, you have 3*2*1 = 6.

$8! = 8*7*6*5*4*3*2*1 = 40320$, which is as close to 25000 as you can get with a factorial.

..placement with repetition

Your original guess was right; the answer is $n^n$.

e.g. for 3 items, you have 3 choices for the first space, 3 choices for the second space, and 3 choices for the final space, so 3*3*3 = 27. For 2 also, i.e. 2*2 = 4 ways ((1|1), (1|2), (2|1), (2|2)).

$6^6 = 46656$ is as close to 25000 as you'll get.

• thanks, as it is so often, before you can know the answer, you need to know the exact question first. I wasn't even aware that there are two possible ways to look at it – Blub Apr 26 '11 at 10:34

A note on terminology and notation...

There are two words you want to learn here, permutations and combinations.

Permutations are how many distinct arrangements can be made from a set, so (1,2,3) and (3,2,1) are different.

$$^nP_r = P^n_r = \frac{n!}{(n-r)!}$$

Combinations are how many subsets you can form from a set, so (1,2,3) is the same as (3,2,1), same elements.

$$^nC_r = C^n_r = \binom{n}{r} = \frac{n!}{r!(n-r)!}$$

Often combinations are spoken as "$n$ choose $r$", as they are the measure of how many choices there are from a set. So the UK National Lottery is a choice of 6 balls from 49, so the number of ways to choose this is "49 choose 6".

$$^{49}C_6 = \binom{49}{6} = 13,983,816$$

You seem to be using the term "variation" in an unorthodox way. An arrangement of $n$ objects is called a permutation. A variation is an arrangement of $k$ objects chosen from $n$ objects. The number of permutations of $n$ objects is $n!$ ("n factorial"), which is defined as the product of all natural numbers up to and including $n$.

He is referring to variation which is arrangement of k-objects from set of n where replacement is possible but order is not important. This is neither permutation nor combination. The answer is C(n+k-1,k) where the symbol C stands for number of combinations.