What is the formula for permutations?

Answer 1

The number of permutations of r objects taken from n distinct objects is nPr = n!/(n-r)!

n! = 1*2*3*... *(n-1)*n

In many cases, one can do permutation problems without the "formula."

Here is a concrete example, then a more abstract version to help see this.

The purpose of pointing this out is not only to make it easier to do this type of problem, but to help understand and remember the formula.

Say we have 10 distinct (we assume distinct from now on) objects and we want to pick 3 objects from the 10. The order of choice is important.

Since there are 10 objects, there are 10 choices when we pick the first one. Now there are 9 left since one is picked so there are 9 choices for the next object. Similarly, there are 8 choices for the third object. The multiplication rule tells us there are 10x9x8=720 ways to pick these three objects.

Now say we have n distinct objects and we want to pick r of them. We have n choices for the first, n-1 choices for the second and n-3 choices for the third.

If n=10 as above and r=3, we saw this was 10x9x8 but in the more general case it is n x (n-1) x(n-2). But how do we know where to stop? We want to pick r items.

So in the case of r=3, the last number in our product was 8. This is (10-3)+1. So in general the last number in the product is (n-r)+1

The general pattern therefore for picking r objects taken from n distinct objects is

n(n-1)(n-2)(n-3)...(n-r+1). This is the same as n!/(n-r)!

The denominator in the formula simply allows us to write n! which is 1*2*3*... *(n-1)*n

and cancel out all the numbers in the product that are less than n-r+1. For example. Using the same example if we pick 3 items from 10 we saw that was 10x9x8. Using the formula we have 10!/(10-3)! or 10!/7! which is (10x9x8x7x6x5x4x3x2x1)/(7x6x5x4x3x2x1) and after cancelling the (7x6x5x4x3x2x1), which occurs in the products in both the numerator and the denominator, we are left with 10x9x8=720 which we know to be the answer.