I will discuss counting two types of picking a group of items from a large number of items. These two types are called *permutations* (also called arrangements) and *combinations* (also called selections).

Combinations are when the order of the picking does not matter. For example, when picking 5 cards from a 52 card deck, the order does not matter: Ace, 2, 3, 4, 5 is the same hand as 5, 4, 3, 2, Ace (assuming the suits are the same). Or another example is how many 5 player teams can be made from 30 people. I will discuss combinations in a subsequent post. This post is about permutations, where the order of things picked does matter.

An example of a permutation problem is how many ways can you arrange 5 guests on a table from a group of 50 people. Here, order matters: Adam, Betty, Charlie, David, Eddie arranged in that order is different from Eddie, David, Charlie, Betty, Adam.

Let’s look at a simple example and extrapolate from that.

From 5 people, how many ways can we seat 3 of them? There are 5 ways to pick the first person. Now there are only 4 people left, so there are 4 ways to pick the next person. Now there are 3 people left so we only have 3 ways to pick the last person. So the number of ways is the 5 ways to pick the first times the 4 ways to pick the second times the 3 ways to pick the last person: 5 × 4 × 3 = 60 ways to arrange 3 people from a group of 5. If you follow this pattern, you can arrange 5 people from a group of 10, 10 × 9 × 8 × 7 × 6 = 30,240 ways.

This can be generalised: how many ways can you arrange *x* things from *n* things. Before I show the formula for this, I need to explain new notation.

Using a “!” after a number has a meaning in maths. This is called a *factorial*. As an example, 5! = 5 × 4 × 3 × 2 × 1 = 120. So a factorial is successively multiplying one less number. Factorials increase quickly. 40! is a number slightly greater than 8 followed by 47 zeroes. Factorials are used in maths formulas frequently and in order to make these consistent, 0! is defined as 1. Doesn’t look right but this must defined this way as we will see.

Looking at the examples above, we have partial factorials: instead of 5 × 4 ×3 × 2 × 1, we have 5 × 4 × 3, or instead of 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 we have 10 × 9 × 8 × 7 × 6. Notice that 5! can be thought of as 5 × 3 × 4 × 2! and 10! can be 10 × 9 × 8 × 7 × 6 × 5!. In the first example, the “2” is 5 – 3, that is the number of people minus the number in the arrangement. In the second, the “5” is 10 – 5, that is the number of people minus the number in the arrangement. If we let *n* be the number of total things and *x* the number of the things to be arranged, then the formula to compute this in general is:

The formula is the far right expression, the notations on the left are the common notations used in different places that mean the same thing. So applying this to our two examples:

\[P(5,3)=\frac{5!}{\left(5-3\right)!}=\frac{5\times4\times3\times2!}{2!}=5\times4\times3=60\] \[P\left(10,5\right)=\frac{10!}{\left(10-5\right)!}=\frac{10\times9\times8\times7\times6\times5!}{5!}=10\times9\times8\times7\times6=30,240\]The 2! and the 5! cancel out in the fractions and we get the result we want. If we wanted to arrange 5 things from a group of 5, we use the definition that 0! = 1:

\[P\left(5,5\right)=\frac{5!}{\left(5-5\right)!}=\frac{5!}{0!}=\frac{5!}{1}=5\times4\times3\times2\times1=120\]Because of how large factorials grow, if calculating this formula by hand, it is better to first cancel the (*n* – *x*)! part from the numerator, then calculate the result.

If you are fortunate to own a CAS calculator, using the permutation function nPr gets the same result with less work: nPr(10,5) = 30,240.

Now this does not directly answer questions about picks where order does not matter, like the number of poker hands. That is a combination question and I will talk about that in my next post.