Permutations follow directly from the fundamental principle of counting. The permutations are all the different number of ways in which we can arrange a number of objects. Since the objects can only be counted from the natural numbers or the counting numbers, we can say that in permutations we will only encounter the positive integers. Here we will introduce the formulae for permutations and see how we can use them to solve the questions of various exams.

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## Permutations

Given a set of N distinct objects, a permutation is an arrangement of the entire set in order without repetitions. There are N! ways to permute the entire set. The value N! is called “N factorial” and is computed by:

N! = N × (N – 1) × (N – 2) × . . . × 1.

This gives the number of permutations for N objects taken all at a time. Suppose a set has N distinct objects and we wish to make a list of ‘k’ of these objects (in order without repeats). For example, from a group of 32 balls, we

need 3 balls for a slot number 1, a slot number 2 and a slot number 3. How many choices are possible?

Answer: We are listing 3 without repeats from a group of 32, so there are 32 × 31 × 30 = 29,760 possible choices. Notice that the number of choices also can be computed by 32!/29! ; but in this case, it is easier to use 32 × 31×30. However, if we were arranging a larger portion of the set, then it would be more convenient to use the factorial notation.

**Browse more Topics under Permutation And Combination**

- Factorial Notation
- Number of Permutations
- Combination
- Number of Combination
- Permutation and Combination Practice Questions

## The General Formula Of Permutations

In the above example, we saw that if we are permuting or arranging 32 objects into 3 slots or in other words 32 objects in three ways, then the number of arrangements can be written as 32!/29!. We can write this as:

32!/(32 – 3)!. If we generalise this, we can see that if we have ‘n’ objects taken say ‘r’ at a time, the total number of permutations is equal to n!/(n-r)!. this is the general formula for permutations. The expression P(n, r), also written ^{n}P_{r}, is calculated by:

P(n, r ) = n!/(n-r)!

You must use the difference in the denominator. For example, P(14, 6) = 14!/(14-6)! = 14!/8! = 2,162,160.

The formula for P(n, r ) gives the number of ways to permute a group of r objects selected from the larger group of n objects.

Remember that in permutations, the order does matter. This means that if we have two letters say A and A, then AA and AA where the order of the two A’s has changed will count as two permutations. Let us see some more examples:

For example, we have two identical balls that we have marked as ‘a’ and ‘a’. Then instead of one arrangement, we count them as two because in permutations the order matters.

Example 1: Suppose you want to arrange your English, Hindi, Mathematics, History, Geography and Science books on a shelf. In how many ways can you do it?

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Answer: Here we have to arrange 6 books. As we know that the number of permutations of n objects is n! = n (n – 1)(n – 2) … 2.1

Here n = 6 and therefore, number of permutations is 6! = 6.5.4.3.2.1 = 720. Therefore the number of ways we can arrange the six books on the shelf = 720.

## Solved Examples For You

Example 2: Suppose you have 6 happy birthday cards for your friends and you want to send them to 4 of your

friends. In how many ways can you send these cards to 4 of your friends?

Answer: Here we have to find the number of permutations of 4 objects out of 6 objects. In other words, we have to count the number of permutations of six objects take 4 at a time. This can be done as follows:

This number is 6(6-1)(6-2)(6-3) = 6.5.4.3 = 360. We can also do this in an easy way as below:

^{6}P_{4} = 6!/(6-4)! = 6!/2! = 6.5.4.3 = 360. Therefore, cards can be sent in 360 ways.

Example 3: In a library, there are 4 books on fairy tales, 5 books are novels and 3 books are on plays. In how many ways can you arrange these so that books on the fairy tales are together in one place. The novels are together and plays are also together. The requirement is that these books should be in a specific order i.e., books on fairy tales, before novels, before plays.

Answer: There are 4 books on fairy tales and they have to be put together. They can be arranged in 4! ways.

Similarly, there are 5 novels. They can be arranged in 5! ways. And there are 3 books on plays.

They can be arranged in 3! ways. So, by the counting principle all of them together can be arranged in 4! × 5! × 3! ways = 17280 ways.

### Type II

Example 4: In the above example what is the number of permutations if the books are not to be kept in order?

Answer: Whenever you are asked to keep a particular class of objects together, a convenient trick is to sort of glue them together in your head and treat them as one object. First, we consider the books on fairy tales, novels and plays as single objects.

These three objects i.e the one group of fairy tale books, the one group of novels and the one group of plays can be arranged in 3! ways = 6 ways.

Let us fix one of these 6 arrangements. This may give us a specific order, say, novels → fairy tales → plays. Given this order, the books on the same subject can be arranged as follows. In other words, now we have to count the internal permutations. The 4 books on fairy tales can be arranged among themselves in 4! = 24 ways.

The 5 novels can be arranged in 5! = 120 ways. The 3 plays can be arranged in 3! = 6 ways.

For a given order, the books can be arranged in 24×120×6 = 17280 ways.

Therefore, for all the 6 possible orders the books can be arranged in 6×17280 = 103680 ways.

## Practice Problems

Q 1: In how many ways can 4 girls and 5 boys be arranged in a row so that all the four girls are together?

A) 568 B) 16850 C) 17280 D) 19874

Ans: C) 17280

Q 2: There are 6 boys who enter a boat with 8 seats, 4 on each side. In how many ways can they sit anywhere on the boat?

A) 20160 B) 30160 C) 40160 D) 50160

Ans: A) 20160