A collection or a combination of objects from a set where the order or the arrangement of the chosen objects does matter is termed as permutation. Hence, we can clearly see that the words in a language are actually some specific permutations of a collection of alphabets taken together. Now let us study more about permutations and its formula in detail below.

A language, English for example, is something that we use to communicate every day. But have you ever wondered what would happen to our understanding of the English language if the order of alphabets which form the words is changed? What if I tell you that a ‘flower’ is spelt as ‘flowre’? Would it make sense to you? No. There you have a very important example of permutation formula used in daily life.

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## Permutation

A permutation is an arrangement of objects in a definite order. Since we have already studied combinations, we can also interpret Permutations as ‘ordered combinations’.

*(Image Source: Wikipedia)*

### Analysis

Let’s say we have a set of ‘n’ distinct objects, out of which we must choose ‘r’ objects. We can proceed in the following two ways of choosing the required objects:

- When the repetition of objects is allowed
- When the repetition of objects is not allowed

### Case 1

When the repetition of objects is allowed, at every step of our choosing an object from the set of ‘n’ objects, we have all the ‘n’ choices available to us since we can make a choice multiple times. So, for choosing ‘r’ objects, we have n choices available to us ‘r’ times. Let us call the event of choosing an object as E:

n(E) = n (the number of ways in which E can take place)

Since this event is taking place ‘r’ times and the act of choosing an object from the available set is always independent of our other choices, we may invoke the Product Rule of Counting here. Using the fundamental principle then, we get,

n(E taking place ‘r’ times) = n^{r}

This is the permutation formula for the number of permutations possible for the choice of ‘r’ objects from a set of ‘n’ distinct objects when repetition is allowed.

### Case 2

In this case, when the repetition of objects is not allowed, we must be careful, not to choose a specific object more than once. Hence our choices after each event get reduced by one. For example, when we begin choosing our first object, we have all the ‘n’ choices available to us.

In the next event, however, we have ‘(n-1)’ objects available for choice, since we must not include the object that we have already chosen in the first step. Similarly, for the third step, we have ‘(n-2)’ objects available to us. Thus, from the Product Rule of Counting, we can get,

n(E) = n × (n – 1) × (n – 1) … (n – (r – 1))

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

= ^{n}P_{r} *(notation for the number of permutations of r objects out of a set of n distinct objects)*

### A Specific Case

The number of permutations of ‘n’ objects from a set of ‘n’ distinct objects would be given by n!. For example, the number of ways in which you can jumble the alphabets of the word ‘flower’ is given by 6! where the number of alphabets in the word are 6. What if our reservoir set of ‘n’ objects has some repeated elements?

The formula can be derived in a similar manner to our derivation of the general formula, but with some important restrictions. You could try working it out yourself! But for the time being, let us just state and understand it. The number of permutations of ‘n’ objects where p_{1 }objects are of one kind, p_{2 }objects are of one other kind… till p_{k }, is given by:

\( \frac{n!}{p1!.p2!…pk!} \)

where clearly, p_{1} + p_{2} +p_{3} …..+ p_{k} = n.

Example: The number of ways in which you can jumble the alphabets of the word ‘balloon’ is given by

\( \frac{7!}{1!.1!.2!.2!.1!} \)

since the number count of different alphabets is given as:

n(occurrence of alphabet ‘b’)= 1

n(occurrence of alphabet ‘a’)= 1

Also, n(occurrence of alphabet ‘l’)= 2

n(occurrence of alphabet ‘o’)= 2

n(occurrence of alphabet ‘n’)= 1

**Browse more Topics under Permutations And Combinations**

## Alternate Analysis of ^{n}P_{r}

If you remember, we had also mentioned ‘ordered combinations’ as another interpretation of ‘permutations’. Let us look at the formula for the number of combinations of ‘r’ objects out of a set of ‘n’ objects:

^{n}C_{r} \( \ = \ \frac{n!}{r!(n-r)!} \)

Let us take a specific combination of these number of different combinations. In permutations, the order of the chosen elements does matter; while in the combinations the order doesn’t matter. Thus, to arrive at the number of permutations, we must multiply the number of combinations by the number of possible permutations of each specific combination!

What is the number of permutations for a set of ‘r’ objects? It is simply r!, as we have discussed some time ago. Multiplying ^{n}C_{r} by r!, we observe:

^{n}C_{r} × r! = \( \frac{n!}{(n-r)!} \) = ^{n}P_{r}

which is the same as the one we derived from our Product Rule of Counting. Therefore, the interpretation of ‘Permutations’ as ‘Ordered Combinations’ is correct!

## Solved Example on Permutation Formula

Question: In a sports broadcasting company, the manager must pick the top three goals of the month, from a list of ten goals. In how many ways can the top three goals be decided?

Solution: Since the manager must decide the top three goals of the month; the order of the goals is very important! It decides the first-place winner, the runner-up, and the second runner-up. Thus, we can see that the problem is of permutation formula.

Picking up three goals from a list of ten:

Possible Permutations = ^{10}P_{3 }= \( \frac{10!}{(10-3)!} \) = 10 × 9 × 8 = 720

Therefore, there are 720 ways of picking the top three goals!

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