# All possible arrangements (Mathematics).

I have a question concerning all possible non duplicative arrangements of a set of things. Dose someone know the formula to calculate this one ?

Lets give a example lets ay I have A-Apple B –Banana O – Orange and I wont to arrange thus items in a order and I would like to know how many possible arrangements there are:

ABO

AOB

BAO

BOA

OAB

OBA

So using a brute force method we get 6 however how can someone calculate something like this using a mathematical formula?

Warning I’m not a native English speaker.

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#1In Discrete mathematics, we count the number of ways to put an object in a position.

In your example, you have 3 objects and three positions, so there are 3x2x1=6

6 ways to arrange it.

You see, you have a number of spaces (in your example three..)

In the first space, you can pick any of the three (Apple, Banana, Orange..) so there are 3 ways to do that.

In the second space t there are only two, since the first slot is occupied, and you used one of the fruits to occupy it so there are 2 ways.

The third slot, you have no choice, it's the one that's left over.

So now you multiply them together.

3x2x1=6

This is used so often it's given a name, the factorial denoted 3!

n!=nx(n-1)x(n-2).......x1

#2Thanks and what would be if I hade more object then spaces to arrange them like 4 object in 3 spaces ?

Warning I’m not a native English speaker.

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#3I believe the formula is:

n! / (k!(n-k)!)

where n is the number of elements(fruit) and k is the number of elements selected (spaces).

Readiness to answer all questions is the infallible sign of stupidity. Saul Bellow, Herzog

#4no. if you had 4 objects and 3 spaces, you would have to use the formula n!/(n-k)! so the result would be 4! (24)

you can check out the wikipedia article on permutations, or just put "permutations" into google and you'll get a lot of resources. you might also be interested in looking up "combinations" as they are related.

#5carx wrote:If you had 5 objects with say 3 spaces than it would be

5x4x3

n!/(n-k)!

#6Sorry, after beating the difference between permutations and combinations into my head for a year I still get it wrong. n over k gives the number of COMBINATIONS without repetition.

Readiness to answer all questions is the infallible sign of stupidity. Saul Bellow, Herzog

#7Jolt wrote:However this one is reserved for numerical systems where N is the base of the system and K is the number of positions.

If you have a lock using decimal numbers and 5 positions all possible combinations including 99999 and 00000 are 100000 possible combinations this works for every system where the “objects” of the set can repeat themselves.

Warning I’m not a native English speaker.

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