Initially, our Rule for Permutations may seem like an odd rule, but the following example may help to make it more intuitive. First recall:

Rule for Permutations
Suppose we have n objects that we want to fill k spots where kn. Then, the number of possible ways ("permutations") to do this is:

    \[\frac{n!}{(n-k)!} .\]

Now consider:

Example
Our refrigerator has 5 magnetic numbers on it: 3, 2, 4, 5, and 7. How many ways are there to order the magnets on our fridge into 5-digit numbers?

Solution
By the question, we have 5 objects to fill into 5 spaces (corresponding to the five digits). Thus, by our formula, we get: 5!/(5-5)! = 5!/0! = 5!/1 = 5! = 120.

But to make this answer more intuitive, consider that we have five spaces to fill in with letters:

__               __               __               __               __               

And for the first space, we can pick any number. It doesn’t matter which one. So we have five options for the first space.

Suppose we picked the number A. (A is just a stand-in for whatever number we actually picked; it doesn’t matter which one we picked). Then, we have:

                                                                                                          

(5 options) 

And now, for the second digit, we can pick any number except for A. So we have four options. Suppose our second digit was B. Then we get:

                                                                                                        

(5 options)  (4 options)

And so on for the rest:

                                                                                                        

(5 options)  (4 options)   (3 options)   (2 options)  (1 option)

Thus, the total number of possibilities we could get is: 5 * 4 * 3 * 2 * 1 = 120, exactly what we got by following our rule.

Now, let’s see how permutations and probabilities can interact:

Example
You are forming four-digit numbers from the numbers 1, 2, 3, and 4. Each of your four-digit numbers must use all of those numbers exactly once. What is the probability that a number randomly chosen from your four-digit numbers contains a prime number in the tens place?

Solution
Now, we want to find:

How many four-digit numbers have a prime number in the tens place?

How many four-digit numbers are there?

To find (1), we want to consider numbers like:

                                (prime)                  

There are only two prime numbers here: 2, 3. Thus, all of our numbers must be like:

                                2                 

or

                                3                 

In either case, we have three choices for the remaining numbers (1, 3, 4 in the first case, and 1, 2, 4 in the second). Thus, there are 6 ways of ordering the remaining three in either case, so we have a total of 12 four-digit numbers with a prime in the tens place.

To answer (2), we want to find how many total four-digit numbers there are. This can be done via our rule: we just need to consider how many ways there are to put 4 objects into four slots. Thus, we get \frac{4!}{(4-4)!} = \frac{4!}{0!} = \frac{4!}{1} = 24 ways to do so.

Thus, the answer is that there is a \frac{12}{24} = \frac{1}{2} chance of getting a prime in the tens place.


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