Difference between Permutations and Combinations

Probability is concerned with the chance or possibility that an event may occur or not occur if there are ‘n’ possibilities. Put simply,  probability tells us the percentage of happening of an event. Probability can be expressed as a number from 0 to1 or as a percentage. 

Event

Event means the outcome of an experiment. For example, when we throw a die (throwing of die is an experiment), any number can be obtained on the top face of the die out of 1,  2, 3, 4, 5, and 6. The appearance of any of these numbers on the die is an event. As stated above probability lies from 0 to 1. An event that is sure to occur has a probability of 1 (100%) and an event that cannot occur at all is called an impossible event and its probability is 0.

Sample Space  

Sample space is the set of all possible outcomes of an experiment. Taking the above example of throwing a die,  the set of all possible outcomes (1, 2, 3, 4, 5, 6) is the sample space.  

Another example is tossing two coins or tossing one coin two times. Here, the sample space is (HH,  HT,  TH, TT). It needs to be understood clearly that the sum of probabilities of all individual events in the sample space is always 1.

Formula of probability  

The most basic formula for calculating probability is,  

P = Number of favorable outcomes of an event / Total number of outcomes in the experiment.  

For example, in the tossing of two coins, we see that the total number of outcomes is 4 out of which a single head appears 2 times (HT, TH). So the probability of getting a single head is  P( getting a single head) = 2/4 = 1/2.

Permutations 

A permutation is a concept that means to arrange a given set of elements in a particular order. Here the sequence of arrangement is important. A simple way to understand permutation is if we have some objects with us and we want to arrange them (It doesn’t matter which object you choose first or last), then in how many ways you can arrange them. Let’s take an example,

If three English alphabets are taken – p, q, and r and we want to arrange them, then these can be arranged like (p, q, r), (p, r, q), (q, p, r), (q, r, p), (r, p, q) and  (r, q, p). Only these six arrangements are possible. Now the word arrangement here is called a Permutation, i.e. only these six permutations are possible.

The formula for finding the number of permutations

If ‘n’ elements are given, out of which we want to arrange ‘r’ elements,  then the number of possible arrangements or permutations is given by,  

ⁿ_rP  = n! / (n – r)!

Look at some examples at the end of this article.  

Combination

The combination is a concept that is concerned with the selection of some elements from a given set of elements. Here the order in which the elements are selected is not important. Now we look at the concept of combination further. The concept of combination implies the selection of some objects out of the given objects. The combination is not concerned with the arrangement of the chosen objects.

For 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.

Formula to find number of combinations

If we have ‘n’ elements out of which we want to select ‘r’ elements then the number of possible combinations is given by  

ⁿ_rC = n! / r!(n – r)!

What is the difference between combinations and permutations?

The definitions of permutation and combination are given above and they are defined in detail. Now let’s take a look at the difference between the two,

Permutations	Combinations
Permutation means total number of ways in which we can arrange elements in The combination order combination.	The combination means a total number of ways in which we can select some elements from a given set of elements.
The order in which the elements are placed is very important.	The order in which the elements are selected is not important.
For example, on tossing three coins, HHT is a different permutation from HTH.	In combination HHT is the same as HTH. The order does not matter.
Permutation can be made using repetition or without repetition of elements.	Combination is not concerned with repetition or without repetition of elements.

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 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 colors 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.