What is the Fundamental Counting Principle? 

The fundamental counting principle is a rule that is used to count the total number of possible outcomes in a given situation. 

If there are "a" ways of performing the first action and "b" ways of performing the second action after the first action, then there are a × b ways to perform both of these actions. 

For example, Sam has 4 pairs of trousers and 3 shirts to wear. 

This means that Sam has, 4 ways of choosing a pair of trousers and three ways of choosing a shirt. For every pair of trousers, there are 3 choices for a shirt. So, there are 4×3=12 ways to choose a pair of trousers and a shirt. 

Let us name the 4 pairs of trousers as P1, P2, P3, and P4 and the 3 shirts as S1, S2, and S3. 

Then, 12 possibilities can be shown as P1S1, P1S2, P1S3, P2S1, P2S2, P2S3, P3S1, P3S2, P3S3, P4S1, P4S2, and P4S3. 

However, if Sam can choose only one pair of trousers or one shirt at any instance, then the number of possibilities will be 3 + 4 = 7. 

The possibilities are: P1, P2, P3, P4, S1, S2, and S3

Permutation 

A permutation is an arrangement of a number of objects in a definite order wherein some or all of the objects are considered. 

For example, the different possible arrangements of letters E, N, O, and S, are NOSE, NEOS, SOEN, EOSN...etc. In this list, each arrangement is different from one another and each arrangement is called a permutation of 4 different letters taken all at a time.  

Now consider another example where we need to determine the number of three-letter words, with or without meaning that can be formed from the 6-letter word 'TYPING', without repeating any letter.  

In this case, the arrangements will be TYP, ING, TYI, YIN, GIN, YPI, ... etc. Here, we cannot consider the arrangement YPY as Y is being repeated. 

So, the permutations of 6 different letters, taken 3 at a time is: 

Factorial of 6 divided by the factorial of 3, that is,

$$6\times 5\times 4=120$$

For the first box, all the 6 letters are available. If we write T in the first box then we are left with 5 letters available for the second box. Similarly, if we write Y in the second box we will be left with only 4 letters available for the third box. This is how we get all the 120 words.

If repetition of the letters is allowed, the number of arrangements of letters will be: 

$$6\times 6\times 6=216$$

This is because every box can have any of the 6 available letters. 

Permutations of Distinct Objects 

The number of permutations of n different objects taken r at a time, where 0 ≤ r ≤ n, and the object does not repeat is: 

$n\left(n-1\right)\left(n-2\right)\mathrm{...}\left(n-r+1\right)$, which is denoted by ${}^{n}P{}_r=\frac{n!}{\left(n-r\right)!}$.

The notation n! or factorial of n represents the product of the first n natural numbers, that is,

$n\times \left(n-1\right)\times \left(n-2\right)\times \mathrm{...}\times 3\times 2\times 1$.

Sine "n" different objects fill the "r" vacant places, so, the first place can be filled in "n" ways, second place in (n – 1) ways, third place in (n – 2) ways, and so on till the r^th place can be filled in (n – (r – 1)) ways.  

Therefore, the number of ways of filling “r” vacant places in succession is

is $n\left(n-1\right)\left(n-2\right)\mathrm{...}\left(n-r+1\right)$.

Let us see how we arrived at the permutation formula ${}^{n}P{}_r=\frac{n!}{\left(n-r\right)!}$, where 0 ≤ r ≤ n.

$${}^{n}P{}_r=n\left(n-1\right)\left(n-2\right)\mathrm{...}\left(n-r+1\right)$$

Multiplying numerator and denominator both by $\left(n-r\right)\times \left(n-r-1\right)\times \left(n-r-2\right)\times \mathrm{...}\times 3\times 2\times 1$, we get:

${}^{n}P{}_r=\frac{n\left(n-1\right)\left(n-2\right)\mathrm{...}\left(n-r+1\right)\left(n-r\right)\times \left(n-r-1\right)\times \left(n-r-2\right)\times \mathrm{...}\times 3\times 2\times 1}{\left(n-r\right)\times \left(n-r-1\right)\times \left(n-r-2\right)\times \mathrm{...}\times 3\times 2\times 1}$

${}^{n}P{}_r=\frac{n!}{\left(n-r\right)!},\text{ where }0\le r\le n$

Sample problem 

If 3 balls are needed to be placed in 2 boxes, in how many ways can the balls be placed? 

This is a permutation problem. The balls can be placed in ³P₂ ways. 

That is ${}^{3}P{}_2=\frac{3!}{\left(3-2\right)!}=\frac{3!}{1!}=\frac{3\times 2\times 1}{1}=6\text{ ways}$.

Permutations of Non-Distinct Objects 

Suppose we need to determine the number of ways of rearranging the letters of the word "APPLE". 

There are two Ps that are similar. If we consider the two Ps to be different from one another, as in P1 and P2, then the number of permutations of all the five letters taken at the same time is 5! 

Now, among the 5! permutation, consider the two outcomes AP1P2LE and AP2P1LE. Corresponding to this outcome, we have 2! permutations AP1P2LE and AP2P1LE because P1 and P2 can be permutated in 2! ways within the word. The two outcomes are different because P1 and P2 are different and not the same letter. However, if P1 and P2 are the same (letter P) at both the positions, then we will end up counting duplicates 2! times. To remove this redundancy, we will have to consider any one outcome and hence divide the total number of permutations by 2! (factorial of 2). 

Therefore, in scenarios having non-distinct or duplicate objects, the required number of permutations

will be $\frac{n!}{p!}$.

Here, n = 5, and p = 2.

So, APPLE can be written in $\frac{5!}{2!}=\frac{5\times 4\times 3\times 2\times 1}{2\times 1}=60$ different ways.

Combination 

The combination is a process of selecting “r” number of items from a given set of “n” number of items, without considering the order of the items as in the fundamental counting principle. 

Let us now assume that there is a group of three badminton players P, Q, and R. A team consisting of 2 players must be formed. In how many ways can this be done? Is the team formed by A and B different from the team formed by B and A? No, they are the same. 

So, there are 3 possible ways in which the team could be constructed. 

These are AB, BC, and AC. 

Here, each selection is a combination of 3 different objects taken 2 at a time or selection of 2 players from 3 players. 

So, we obtain the formula for finding the number of combinations of n different objects taken r at a time, denoted by ${}^{n}C{}_r$where,

$${}^n{C}_r=\frac{{}^{n}P{}_r}{r!}=\frac{n!}{\left(n-r\right)!r!}$$

Sample problem 

A bag contains 5 red, 4 purple, and 3 green balls. How many different combinations of 2 red, 2 purple, and 1 green colored ball can be selected? 

Here order is not important. So, we need to count the combinations. 

Therefore, there are ⁵C₂ ways of choosing 2 red balls from 5 red balls, ⁴C₂ ways of choosing 2 purple balls from 4 purple balls, and ³C₁ ways of choosing 1 green ball from 3 green balls. 

So, the total number of ways of selecting the required balls = ${}^{5}C{}_2\times {}^{4}C{}_2\times {}^{3}C{}_1=60$.

Difference between Permutations and Combinations 

Here are some of the main differences between permutation and combination. 

Formulas 

Common Mistakes 

Some of the possible common errors when dealing with permutation and combination are as follows: 

• Understanding the question incorrectly– not understanding whether repetition is allowed or, whether there are some other constraints for arrangements and selection. 
• Using improper denominator – we may use an incorrect r! or use an incorrect (n-r)!. 
• Making errors in calculations. 

Context and Applications 

A permutation helps to arrange several items and find the total number of required arrangements without writing down all the possibilities, which is very time-consuming. Permutations will help to arrange fruits and juices in a shop, books in a library, students in a queue, etc.

Similarly, a combination helps to select few items from many given items and to determine the number of ways in which we can select such items without writing down all the available options. It helps to form a committee from all the members, and choose the captain and vice-captain of a team from all players, favorite foods from a given menu, colors from a color palette, clothes from a wardrobe, etc.

The above topic is studied in the following courses.

• Bachelor in Mathematics
• Masters in Mathematics