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## Permutation and Combination and What is a Factorial? If you have 7 different options on the menu and you can order 2 of them for your combo-meal, how many possible meal combinations are there?

To figure this out, you need only a simple formula and a cool math operation called the factorial.

### What is a factorial?

The factorial is written as an integer with an exclamation point (!).  This symbolizes that all integers from the one given, all the way down to 1, are to be multiplied.  For example, 5! (said “five factorial”) is 5 x 4 x 3 x 2 x 1 which equals 120.  5! is equal to 5 x 4 x 3!, since 3! is 3 x 2 x 1.  Another property of the factorial is that 0! is equal to 1, for no other reason really other than it must be defined that way to make equations using factorials work properly.

### How to calculate combinations

With the factorial, we can use the formula for computing combinations.  The formula uses two variables, n and k, and n is the total number of the objects to select from and k is the number of those objects to be chosen in each selection. $\frac{n!}{k!(n-k)!} = \text{C(n,k)}$

#### Example

If a standard deck of 52 cards is used to deal a player 5 cards, how many possible hands are there?

C(n,k) = C(52,5) = 52!/5!(52-5)! = 52!/5!(47!) = (52 x 51 x 50 x 49 x 48 x 47!)/5!(47!) = (52 x 51 x 50 x 49 x 48)/5! = 2,598,960

### Permutation and Combination and Combination and Permutation

A combination is any unique selection (or subgroups) from the group, where a permutation considers the order of those selections.  We can use the letters in the word MATH as an example. If we were to find all of the different combinations of 2 letters, we would have 6 possibilities:

MA,MT,MH,AT,AH,TH

In this case, the MA is the same as AM – the order does not matter, only the specific members of the selection.  If we wanted to identify how many unique subsets of size 2, where each order matters, we would have the following list of 12 values (the number of permutations of the letters of MATH, taken 2 at a time):

MA,MT,MH,AT,AH,TH,AM,TM,HM,TA,HA,HT

### How to calculate permutations $\frac{n!}{(n-k)!} = \text{P(n,k)}$

Notice the similarity to the combination formula?  Without the additional factor in the denominator, the permutation calculation will result in larger values than the combinations.  This is logical, since the combination subsets can be reordered to create new permutations, which means there are additional possibilities to be counted.

#### Examples

For the hand of 5 cards from the above example, how many ways can it be reordered to create unique permutations?

P(n,k) = P(5,5) = 5!/(5-5)! = 5!/0! = 5!/1 = 5! = 120

And how many separate sequences of 5 cards from the deck of 52?

P(n,k) = P(52,5) = 52!/(52-5)! = (52 x 51 x 50 x 49 x 48 x 47!)/(47!) = 52 x 51 x 50 x 49 x 48 = 311,875,200