Difference between Permutation and Combination

Permutations and combinations are key concepts in mathematics used for counting arrangements and selections. The main difference is that order matters in permutations, while it does not matter in combinations.

Permutation: Permutation is a concept that means to arrange a given set of elements in a particular order.

Example: If we take three English alphabets — p, q, and r and arrange them, the possible orders are:
(p, q, r), (p, r, q), (q, p, r), (q, r, p), (r, p, q), and (r, q, p).

There are only six different arrangements. Each arrangement is called a permutation, because the order of letters matters.

Combination: A Combination is a concept that is concerned with the selection of some elements from a given set of elements. 

Example: The selection of 11 players from a wide number of players for a cricket team comes under combination (that’s it, only selection) but which player will bat first, which will bat second, and so on, this arrangement of players comes under permutation.

The key differences are shown in the table below:

Note: For given values of n and r, the value of the permutation is always greater than the value of the combination (when r > 1). This is because permutations count different arrangements where order matters, while combinations count only selections where order does not matter. Hence, permutation is always greater than combination.

Related Articles:

• Permutation and Combination
• Permutation and Combination - Aptitude Questions
• Quiz about Permuation and Combination

Sample Problems

Question 1: In how many ways can you arrange the letters of the word ARTICLE, taking 4 letters at a time, without repetition, to form words with or without meaning?
Solution: 

Here from 7 letters of the word ARTICLE, we have to arrange any 4 letters to form different words.
So, n = 7 and r = 4. 

Using permutation formula ⁿ_rP = n! / (n - r)!    

⁴₇P = 7! / (7 - 4)! 
= 7!/3! 
= (7 × 6 × 5 × 4 × 3!) / 3! = 7 × 6 × 5 × 4 = 840

Thus there are 840 different ways in which we can arrange 4 letters out of the 7 letters of ARTICLE to form different words.

Question 2: How many 6 digit pin codes can be formed from the digits 0 to 9 if each pin code starts with 48 and no digit is repeated?
Solution:

Here arrange 6 digits from 0 to 9 but the first two digits of the pin code has been already decided (4 and 8). 

So we have to now arrange only 4 digits out of the remaining 8 digits (0, 1, 2, 3, 5, 6, 7, 9).
So, n = 8 and r = 4,  

⁸₄P = 8! / (8 - 4)! 
= 8! / 4! 
= (8 × 7 × 6 × 5 × 4!) / 4! 
= 8 × 7 × 6 × 5 
= 1680

Thus, 1680 different permutation in which 6 digit pin codes can be formed.

Question 3: Out of 10 students, 4 are to be selected for a trip. In how many ways can the selection be made?
Solution:

In this question select 4 students out of given 10. So combination will be used here to find the answer.
n = 10 and r = 4,  

¹⁰₄C = 10! / 4!(10 - 4)! 
= 10! / 4!6! 
= (10 × 9 × 8 × 7 × 6!) / (4 × 3 × 2 × 1 × 6!) 
= (10 × 9 × 8 × 7)/(4 × 3 × 2 × 1) 
= 210

Thus there are 210 different ways of selecting 4 students out of 10.

Question 4: A bag contains 3 red, 5 black, and 4 blue balls. How many ways are there to take out three balls so that each of the colours is taken out?
Solution:

Here take out three balls of each colour. The order in which the balls are taken out does not matter. So use combination to find the answer.

Number of ways of selecting one red ball out of 3 red balls = ³₁C
Number of ways of selecting one black ball out of 5 back balls = ⁵₁C  
Number of ways of selecting one blue ball out of 4 blue balls = ⁴₁C
Total number of ways of selecting three balls of each colour = ³₁C × ⁵₁C × ⁴₁C                 

= 3 × 5 × 4 
= 60

Thus there are 60 ways of selecting three balls of each colour.