## FANDOM

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Permutations are the number of ways one can place items where order is important. For example, if the objects to be ordered were people standing in a queue, it would be the number of ways the people could line up where it matters who is first, second, third, etc. This differs from combinations as combinations don't worry about order.

The simplest permutations occur when all items being ordered are used in the ordering process, for example if 6 people could stand in a queue and all 6 of them do stand in it. To calculate this, you simply need to use factorials, as the number of items which can be placed in the first position is $n$ and the number which can be placed in the second $(n-1)!$ and the third $(n-2)!$, down until there is only one place left for the last item. To find the number of ways n items could be ordered, simply use $n!$. In the queue example above, this would be $6!=720$.

A variation on this idea occurs when there are multiples of the same item which can be ordered. Let's say we have to find how many ways the letters in the word "ADDITION" could be ordered. One can see that in the word addition, there are two Ds and two Is. To factor this into the equation, we need to divide the factorial of all letters by the factorials of the number of repetitions of each similar object, which in this example means the equation would be written $8!/2!2!=10080$ as the total number of letters is 8 and the number of repetitions of D and I are both 2. In a more general statement, a group of n objects where one item has p duplicates including itself and another item has q duplicates including itself and another r duplicates including itself, the equation to find the number of permutations is $n!/(p!q!r!)$.

However, more complex permutations don't use all available items, in which case we use a different formula. When ordering r items out of a possible n items, we say we have a "permutation of n objects taken r at a time". This can also be written as $^nP_r$. To calculate this, we use: $n!/(n-r)!$. For example, let's say that only 4 of the 6 people who could stand in our queue actually do. We can express this as $^6P_4$, with the number of possible permutations being $6!/(6-4)!=6!/(2)!=720/2=360$, so there are 360 ways they could be ordered.

When the items being ordered are placed in a circle, we add yet more complications into the equation. As it doesn't matter where a circle "starts", we remove n ways from the simple factorial equation. In other words, if there are n items to be ordered, the equation to find the number of ways they could be ordered is $n!/n=(n-1)!$. If the items being ordered are able to be viewed from the opposite side, such as in the case of beads on a necklace, this equation is again halved as one does not need to consider the mirrored patterns because they are the same as the originals viewed from the other side. So the number of ways this can be expressed is $(n-1)!/2$.