Once you’ve figured out whether to use permutation or combination, there is actually very little work to be done if you know the formulas for permutation and combination or if you know where the function is hidden on your calculator.
As I mentioned before, permutation is used when the order matters and combination when you just want to choose, but not order, the items. Let’s dive straight into the formulas then go through a few examples.
The number of ways of permuting r objects out of n objects is given by
n P r =n!/(n-r)!  where n! = n * (n-1) * (n-2) * … * (2) * (1)
The number of ways of combining r objects out of n objects is given by
n C r = n!/r!(n-r)!
Note that 0! is defined as 1.
Example 1
How many ways are there of arranging 4 letters out of the following F R I E N D?
FRIEND has 6 different letters, and since the order of letters matter, I know I have to use the permutation formula. Applying the formula directly, the answer is given by
6 P 4 = 6!/(6-4)! – 6!/2! – 360
You can double-check this by using the fundamental counting principle that we covered in Part I. Drawing a tree diagram will also work, but it might get a little messy as the tree gets bigger and bigger.
Example 2
If there are 3 entrees and 5 desserts, how many ways are there of choosing 1 entrée and 2 desserts? Note, that the question does not say anything about the order in which the entrees and desserts are eaten, so we know to use the combination principle. Because we are choosing entrees and desserts separately, we have to apply the combination formula twice.
First, to choose 1 entrée out of 3, we apply the formula 3 C 1 = 3!/1!(3-1)! – 3!/1!2! – 3
Next, to choose 2 desserts out of 5, we apply the formula 5 C 2 = 5!/2!(5-2)! – 5!/2!3! – 10
Then because for each entrée, there are 10 possible 2-dessert combinations and there are 3 ways of choosing 1 entrée, to get the total number of possibilities, we take 3*10 = 30.
Example 3
What if the question throws you a curveball and asks you to permute something that has a repeated item. For example, how many ways are there of arranging the letters A G H A S T?
The ‘A’ is repeated twice so first we pretend that the letters are distinct and find the number of possibilities. Then divide that value by 2! because ‘A’ is repeated twice.
AGHAST has 6 letters, and if we permute all the letters, we get 6!
Because A is repeated, we divide 6! by 2! to get the answer 360
Supposing I asked you to permute only 4 out of 6 of the letters from AGHAST, then you would do 6 P 4 as per normal, and divide that answer by 2! since A is repeated. The answer should be 180.
Please visit the Grockit forum or leave a comment here to discuss further.
