Permutations

When we deal with probability, we can deal with many things when ordering objects or trying to calculate the probability of a certain event(s). We begin here with Permutations.

A permutation is an ordered arrangement of an object(s). The simplest example of a permutation is arrangements of letters.

(a)  
Write out the arrangements of the letters a,b,c.

(b)  
How many arrangements are there?

Solutions:

(a):

a b c
a c b
b a c
b c a
c a b
c b a

(b):

There are a total of 6 arrangements.

Now let's examine how this came to be. Imagine that you are starting from the beginning. There are 3 letters to choose from (a, b or c) so you write one down. Next, there are still 2 letters to choose from (say b or c) so you pick one and write it down. Finally, there is just the one letter to choose from. Here is how it looks written down:

3 x 2 x 1 = 6

This is the same as saying 3! (3 factorial). This will work the same for any number of objects. Say you had 4 letters to work with. The number of possible arrangements would be 4! or 24. The same for 5, 6, 7, ... n number of objects.

(a)
Six books are on a shelf.  In how many ways can you arrange them?

(b)
Using the letters t, u, v, w, x, y and z, how many permutations 
of those permutations have the letters "ty" next to each other 
in the beginning?

Solutions:

(a):

There are 6! ways of doing these arrangements or 720 total ways.

(b):

To write this out, you are trying to determine the number of permutations when the letters "ty" appear in the beginning:

t  y  __  __  __  __  __

There are only 5 letters remaining to fill the empty spots, so there are 5! ways of doing so or 120 total ways.

Select Permutations

We have seen in the above the number of ways of arranging all possible items. But say we only want to order a select few of those items. There is a way to do so by this formula:

n P r

where n is the total number of objects and r is the number of items you are selecting.

So lets say that you only want to select 3 letters from a selection of 6 in total. This is now seen below:

6 P 3  =  6 * 5 * 4  =  120

So there are 120 ways of doing so. There is also a more clever way of writing the above since at some point you may be dealing with large numbers and do not want to write a long set of products as you would above:

So using the above example, we get the following:

It is seen above that the 3 x 2 x 1 will cancel to just be left with the 6 x 5 x 4 to get the correct answer of 120.

(a)  
Evaluate 10 P 2

(b)  
Evaluate 9 P 5

(c)
Evaluate 4 P 4

(d)
Evaluate 7 P 0

Solutions:

(a)

(b)

(c)

(d)

Distinguishable Permutations

A permutation of objects can be distinguishable if a permutation of an object(s) cannot be obtained from another simply by rearranging the positions of the elements of the same kind. For example, say we had 10 disks (6 red and 4 of other colors) how many distinguishable permutations are there? The key point here is to note that the 6 red cannot be distinguished when rearragning them.

This leads to the general formula:

where n is the total number of elements, n1 are objects of the first kind, n2 are objects of the second kind, n3 are objects of the third kind etc...

Using the above example, we get:

When using this forumla, we must take into account the number of objects that need to be distinguished; here it was 6 red disks. The other 4 do not count in this formula.

Find the number of ways of placing 15 disks in a row given that 
4 are red, 6 are black, 2 are green and 3 are yellow.

Here we take note that there needs to be distinguishable permutations so this leads to the formula:

Challenges

After viewing this tutorial, you should be able to solve the following challenges:

Probability 1
Probability 2
Probability 5
Probability 6

Good luck! Remember to register for a FREE account today to submit the answers to the challenges!