Permutations: Distinct Objects

Introduction 

A permutation is an arrangement of objects in a definite manner. It becomes challenging to know the number of possible arrangements if the number of objects is more. Like, if we were to find out how many ways 1,2,3,4,5,6,7,8,9 can be arranged, it would be not easy to tell by manually putting them in a different order. Permutation helps us calculate the many possibilities of arrangements in such cases.

Permutation formula 

The formula provides an easy way to find the total number of arrangements:

n! = n(n-1) (n-2) ………………..3*2*1

nPr = n!/(n-r)! = n(n-1) (n-2)…(n-r+1)   (valid only for n≥r )

(i.e. the number of permutations of n different objects taken r at a time) 

This formula can also be used to create further the permutation calculator where there are different factors involved.

Theorem 1: where 0 < r ≤ n and the objects do not repeat, the formula then becomes: 

nPr  = n (n – 1) (n – 2)…..(n–r+1) or  

     = n!/(n-r)!

Theorem 2: The number of permutations of n objects, where p1 objects are of one kind, p2 are of the second kind, …, pk is of kth type and the rest, if any, are of a different kind, the formula then becomes: 

nPr = n! / ( p1! * p2! * p3!…..pk!)

Permutation examples

Q: How many ways could these numbers be arranged (1,2,3,4)?

Solution: 

4*3*2*1 = 24

There are 24 ways of arranging the numbers 1,2,3,4.   

Q: Find the total number of arrangements in the following permutations 

(i) 5! (ii) 7!/5! (iii) 8P5

 Solution:

(i) 5! = 5*4*3*2*1 = 120

(ii) 7!/5! = (7*6*5*4*3*2*1) / (5*4*3*2*1) = 42

(iii) 8P5 = 8! / (8-5)! = 8! / 3! = (8*7*6*5*4*3*2*1) / 3*2*1 = 6720

Examples of permutations with the distinct objects

 Q: Consider the word “CARD”, which has distinct objects C, A, R,& D.

How many ways could four and two-letter words be formed using the letters in CARD when letters are not allowed to be repeated? Also find the number of 4 letter words if repetition is allowed?

Solution: Given that we need to create four-letter words, let’s assume that we have four brackets.

Scenario 1: 4 letters, repetition is not allowed.

In this scenario, bracket one can have any of the four letters “C, A, R, D” = 4 choices

bracket two can have any of the remaining three letters as one letter is now inside bracket 1 = 3 choices

bracket three will can any of the remaining two letters = 2 choices

bracket four will have only one choice: the remaining letter = 1 choice

If you multiple all these choices: 4* 3 * 2 *1, you will get the number of possible four-letter words with “C A R D” without repetition.

Thus the number of 4 possible letter words are 4* 3 * 2 *1 = 24 words when repetition is not allowed. 

Scenario 2: 4 letters, repetition is allowed.

In this scenario, bracket one can have any of the 4 letters “C, A, R, D” = 4 choices

bracket two can also have any of the four letters given repetition is allowed = 4 choices

bracket three can also have any of the four letters given repetition is allowed = 4 choices

bracket four can also have any of the four letters given repetition is allowed = 4 choices

If all these choices are multiplied: 4*4*4*4, we get the number of possible four-letter words with “C A R D”.

Thus the number of possible four-letter words are 4*4*4*4 = 256 words, given repetition is allowed.

Scenario 3: 2 letters, repetition is not allowed.

In this scenario, bracket one can have any of the four letters “C, A, R, D” = 4 choices

bracket two can have any of the remaining three letters as one letter is now inside bracket 1 = 3 choices.

There is no bracket 3 & 4, as we have to find two-letter words (CA, CD, RD, RA, etc.)

Now, multiple all these choices: 4* 3, we will get the number of possible two-letter words with “C A R D,” without repetition.

Thus the number of possible two-letter words is 4*3 = 12 words, given repetition is not allowed.

Scenario 4: 2 letters, repetition is allowed.

In this scenario, bracket one can have any of the four letters “C, A, R, D” = 4 choices

bracket two can have any of the four letters given repetition is allowed = 4 choices

There is no bracket three and four, as we have to find two-letter words (like CC, AA, CA, CD, RD, RA etc.)

Now, multiplying all these choices: 4* 4 , we will get the number of possible two-letter words with “C A R D”.

Thus, the number of possible two-letter words are 4*4 = 16 words, given repetition is allowed.

Thus, we have seen a total of 4 scenarios:

1. The number of arrangements of complete set objects without repetition.
2. The number of arrangements of complete set objects with repetition.
3. The number of arrangements, smaller set objects without repetition.
4. The number of arrangements, smaller set objects with repetition.

Conclusion

The Different formulas that can help with Permutations are:  

General Formulas: 

n! = n(n-1) (n-2) ………………..3*2*1

nPr = n!/(n-r)! = n(n-1) (n-2)…(n-r+1) for n>r

Permutations in different Scenarios:  

Scenario 1: Without repetition and permutation of whole set

Permutation = n * (n-1) * (n-2)…1

Scenario 2: With repetition and permutation of the whole set

Permutation = n* n * n …n times

Scenario 3: Without repetition and permutation of subset with r elements

Permutation = n * (n-1) * (n-2)…(n-r+1) 

Scenario 4: With repetition and permutation of the subset with r elements

Permutation = n*n*n…r times