Permutations and combinations tell us different ways in which we can select the objects from a set. Generally, when we solve permutations and combination problems, it is given that the items will be chosen without replacement.

You may be wondering that if both the combinations and permutations are doing the same job, then why different terminologies are used to describe them?

Well, though they perform the same job, however, there is a fundamental difference between these two terminologies. The combination is the selection of items from the set when order is not important. On the other hand, permutation is the selection of items from a group when the order is important. In other words, we can say that permutation is an ordered form of combination.

Let us explain the difference between two terminologies through the following instances:

  1. This basket has carrots, onions, and potatoes.
  2. The pin code of my cell phone is 8745.

In the first example, we do not care about the order of three vegetables. We only care about the types of vegetables that are in the basket. Therefore, it is an example of a combination.

In the second example, we have four digits. The pin code will work in the same order exactly. If the order of the digits is changed, then it will not work. Hence, it is an example of permutation.

 

Permutations Formula

The formula for permutations is given below:

P (n, r) = \frac {n!}{n - r!}

 

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Types of Permutations

There are two types of permutations:

  1. Permutations with repetitions
  2. Permutations with no repetition

Permutations with Repetitions

To compute the permutations when repetition is allowed is straightforward. For example, consider the following scenario:

There are "n" different types of objects. It means that we have a choice to select from  the"n" number of items. Suppose we are asked to select 4 of the objects from the "n" number of objects. The permutations will be:

n x n x n x n

It means that we will multiply n four times.

Now, let suppose we have to choose m number of items from a collection of n different types. In this case, we will multiply "n" "m" times like this:

n x n x n x n x n x ......(m times)

We can easily write this permutation by employing the exponent of m like this:

n ^ m

Consider another scenario below:

We select two numbers from the set of the first 100 natural numbers (1, 2, 3, ......100) in such a way that the repetition is allowed. We will write the permutations as:

100 \cdot 100 = 100^2 = 10000

We will get 10,000 permutations.

Hence, we can conclude that the formula for the possible permutations when repetition is allowed is:

n ^m

Here:

reflects the number of objects to select from

represents the number of objects we want to select

 

Permutations with no repetition

While calculating the permutations when repetition is not allowed, we have to reduce the number of available alternatives each time we select an item. Consider the following example to understand it:
Hannah is asked to choose 7 balls from the pool. All balls have different colors.
In the above example, when Hannah will choose, let say, a red ball, she will not be picking it up again the next time. The number of choices now has been reduced to 6.
Let suppose the next time she picks up a green ball. She cannot choose the green ball again because she has already availed this choice. Hence, the number of choices has reduced further to 5 balls. The choices will continue to reduce until she picks up the last ball.
Now, let suppose she chooses four balls only. What will be the permutations in this case?
Well, if she selects 3 balls from the set of 8 balls, then the permutations will be calculated like this:
8 x 7 x 6 = 336
It means that there are 336 various ways of picking up 3 balls that can be arranged out of 8 balls.

If all the elements from the set needs to be used or selected then we calculate the permutations by using the formula:

P_n = n!

n! is read as n factorial and it is computed by multiplying the numbers from 1 to n together.

Let us consider that in the above example Hannah selects all the pool balls. The number of permutations will be:

8! = 40320

We got the number 40320, like this:

8! = 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 40320

The formula for computing the permutations when repetition is allowed is given below:

P(n,r) = \frac{n!} {n - r!}

Here, n is the number of objects to select from

r represents the number of items we want to select

Now, let us solve some of the examples of permutations.

 

Example 1.

Calculate P (6,4).

Solution

The formula for computing the permutations is given below:

P(n,r) = \frac{n!} {n - r!}

Substitute n = 6 and r = 3 in the above formula to get the result:

P(6,4) = \frac{6!} {6 - 4!}

P(6,4) = \frac{720} {2}

= 360

 

Example 2

Calculate P_5.

Solution

In this case, we don't have the value for r. We only have the value of n. Hence, we can assume that here we will use the formula of permutations when repetition is allowed.

P_5 = 5! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 120

 

Example 3

Suppose there is a set of 7 chocolates. In how many ways, 4 chocolates can be selected from this set (without repetition)?

Solution

In this example, we are given the following information:

The total number of chocolates = 7

Number of chocolates we want to select = 4

Is repetition allowed = No

The formula for calculating the permutations when no repetition is allowed is given below:

P (n, r) = \frac {n!}{n - r!}

Substitute the values in the above formula to get the number of ways the chocolates can be selected:

P (7, 4) = \frac {7!} {7 - 4!}

= \frac {5040} {6} = 840

Hence, we can select 4 chocolates from a group of 7 chocolates in 840 ways.

 

 

Example 4

Suppose there is a set of 9 balls. In how many ways, 5 balls can be selected from this set (without repetition)?

Solution

In this example, we are given the following information:

The total number of balls = 9

Number of balls we want to select = 5

Is repetition allowed = No

The formula for calculating the permutations when no repetition is allowed is given below:

P (n, r) = \frac {n!}{n - r!}

Substitute the values in the above formula to calculate the number of ways the chocolates can be selected:

P (9, 5) = \frac {9!} {9 - 5!}

= \frac {362880} {24} = 15120

Hence, we can select 5 balls from a set of 9 balls in 15, 120 ways.

 

Example 5

Suppose there are 4 girls and 3 boys in a classroom. The teacher wants a seating arrangement such that the girls occupy the even places in a row. How many seating arrangements are possible?

Solution

We are given the following information in this problem:

The number of positions = 7

The even positions for girls will be 2, 4, 6, and 8.

The girls can occupy these 4 even places in 4 ways because there are 4 girls:

P_4 = 4!

P_4 = 24

There are 3 positions left which will be occupied by the boys.

P_3 = 3!

P_3 = 6

Use the counting principle to compute the number of possible arrangements.

Hence, the total number of seating arrangements possible are:

24 \cdot 6 = 144

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Rafia Shabbir