The term 'permutations and combinations' both have different meanings. Permutation is an arrangement of items, where its order matters. For instance, take any three names of your friends as follows: Martin $(M)$, Victor $(V)$ and Bob $(B)$. We can arrange these names in many different orders. We can rearrange them and then find out the various possible outcomes as follows: They will be like $MVB, VMB, VBM, BVM, BMV$ and $MBV$. So, there are 6 ways to arrange three people. The number of count can also be calculated by using the formula: $n!$ = $(n)\ (n  1)\ (n  2)$ . . . . (Sign ‘$!$’ is factorial) Here, in the above example the ‘$n$’ is three. Therefore, $3!$ = $(3)\ (2)\ (1)$ = $6$ The formula for calculating permutation is $P(n, r)$ = $\frac{n!}{(n  r)!}$ Combination is a collection of items, where its order does not matter in a different arrangement. Now, we will consider the example of Martin, Victor and Bob. But, in the combination process, the order does not matters. So, there is no need of arrangement of their names. Now, let us consider another example to understand it more accurately. Take the example of five persons, where we have to arrange them in the group of three. So, to get the arrangement, we will use the combination formula which is stated below: $C\ (n, r)$ = $\frac{n!}{r!\ (n  r)!}$ Now, we will use this however the order does not matter. $C\ (5, 3)$ = $\frac{5!}{3!\ (5  3)!}$ = $\frac{120}{12}$ = $10$.

Each of the different arrangements which can be made by taking some or all of a number of things is called a permutation. The number of permutations of n different objects taken '$r$' at a time is denoted by $^nP_r$ or $P(n,\ r)$ and its value is given by $n\ (n  1) (n  2).......\ (n  r + 1)$, when the specification of object is not allowed.
$^nP_r$ = $n\ (n  1)\ (n  2)\ ....\ (n  r + 1)$
= $\frac{n!}{(n  r)!}$
If repetition of object is allowed, then $^nP_r$ = $n^{r}$
Solved Example
Question: Find the number of permutation, where value of ‘n’ is 7 and the value of ‘r’ is 4.
Solution:
Solution:
We know that formula for permutation is given by:
Permutation _{} $^nP_r$ = $\frac{n!}{(nr)!}$
Given, the value of n = 7 and value of r = 4.
First, we can find the factorial of 7.
The factorial of 7 is 7 x 6 x 5 x 4 x 3 x 2 x 1 = 5040.
Now, we find the factorial of (7  4).
The factorial of (7  4) is (7  4)! = 3!
So, the factorial of 3 is 3 x 2 x 1 = 6
Now, divide 5040 by 6
Permutation = $\frac{5040}{6}$ = 840.
Permutation _{} $^nP_r$ = $\frac{n!}{(nr)!}$
Given, the value of n = 7 and value of r = 4.
First, we can find the factorial of 7.
The factorial of 7 is 7 x 6 x 5 x 4 x 3 x 2 x 1 = 5040.
Now, we find the factorial of (7  4).
The factorial of (7  4) is (7  4)! = 3!
So, the factorial of 3 is 3 x 2 x 1 = 6
Now, divide 5040 by 6
Permutation = $\frac{5040}{6}$ = 840.
$^nP_r$ = $\frac{n!}{(nr)!}$
Where, the values of ‘$n$’ and ‘$r$’ are the non negative integers. Also, the value of ‘$r$’ is less than or equal to ‘$n$’.
‘$r$’ denotes the size of each permutation.
‘$n$’ denotes the size of the set from which element are permuted.
‘$!$’ denotes the factorial operator.
Each of the groups or selections that can be made by taking some or all of a number of things is called a combination. Thus, the selections that can be made by taking the letters $a,\ b,\ c$ two at $a$ time are $ab,\ bc$ and $ac$. These three groups or selections are called the combination of three things taken $2$ at a time. The number of combinations of $n$ dissimilar things taken '$r$' at a time is $^nC_r$.
In combination, order does not matters or we can say it is an unordered collection of a unique size. Now, we will define combination with the help of formulas.
Formula to find the combination is given as follows:Combination = $^nC_r$ = $\frac{n!}{r! (n – r)!}$
A factorial of a number $n$ is the product of all numbers from $1$ to $n$.
$C\ (n, k)$ = $\frac{n!}{k! (n – k)!}$
$C\ (n, k)$ = $\frac{n \times (n – 1) \times … (n – k + 1) \times (n – k) \times (n – k – 1) \times … \times 2 \times 1}{k! (n – k) \times (n – k – 1) \times … \times 2 \times 1}$
$C\ (n, k)$ = $\frac{n \times (n – 1) \times … (n – k + 1)}{(k!)}$
Let us understand combination with the help of an example.
Solved Example
Question: Suppose value of ‘n’ is 8 and value of ‘r’ is 5. Then, find the combination for the given value.
Solution:
Solution:
Formulas to find Combination is as follows:
Combination = $^nC_r$ = $\frac{n!}{r! (n  r)!}$
Given, n = 8 and r = 5.
$^8C_5$ = $\frac{8!}{5! (8  5)!}$
Combination = 56.
Combination = $^nC_r$ = $\frac{n!}{r! (n  r)!}$
Given, n = 8 and r = 5.
$^8C_5$ = $\frac{8!}{5! (8  5)!}$
Combination = 56.
$^nC_r$ = $\frac{n!}{r! (n – r)!}$
where, the values of ‘$n$’ and ‘$r$’ denote nonnegative integers and value of ‘$r$’ is smaller than or equal to value of ‘$n$’. ‘$r$’ shows the size of each combination, ‘$n$’ shows the size of the set and ‘$!$’ denotes the factorial operator. Given below are some of the example problems on combinations.
Solved Examples
Question 1: Find the number of ways in which 5 players out of 8 players can be selected to form a team.
Solution:
Solution:
The required number of ways is $^8C_5$ = 56 ways.
Question 2: Out of 4 men and 3 women, a committee of 3 members is to be formed so that, it has 1 woman and 2 men. In how many different ways can this be done?
Solution:
Solution:
Selection of 1 woman from 3 women can be done in $^3C_1$ ways and selection of 2 men from 4 men can be done in $^4C_2$ ways.
By fundamental principle, these two can be done in ($^3C_1$ x $^4C_2$) = 18 ways.
By fundamental principle, these two can be done in ($^3C_1$ x $^4C_2$) = 18 ways.
Solved Examples
Question 1: How many 4 digit numbers can be made from digits 1, 2, 3, 4, 5, 6 with no two digits common or repeated?
Solution:
Solution:
Required number is equal to permutations of 6 different digits taken 4 at a time.
$^6P_4$ = $\frac{6!}{(64)!}$
= $\frac{6!}{2!}$
= 360
Therefore, using four digit numbers, 360 numbers can be made with no two digits repeated.
$^6P_4$ = $\frac{6!}{(64)!}$
= $\frac{6!}{2!}$
= 360
Therefore, using four digit numbers, 360 numbers can be made with no two digits repeated.
Question 2: There are 4 candidates for the post of a lecturer in Mathematics and
one is to be selected by votes of 5 men. Find the number of ways in which the
votes can be given.
Solution:
Solution:
Each man can vote for one of the 4 candidates and this can be done in 4 ways. Similar is the case with every other man, as repetition is allowed.
5 men can vote in $4^{5}$ = 1024 ways.
Therefore, the votes can be given in 1024 ways.
5 men can vote in $4^{5}$ = 1024 ways.
Therefore, the votes can be given in 1024 ways.
Solved Examples
Question 1: In how many ways can 3 people be seated in a row containing 6 seats?
Solution:
Solution:
First person can be seated in 6 ways, the second person in 5 ways and the third person in 4 ways.
Then, by fundamental principle, total no of ways on which 3 person can be seated in 6 seats in a row is (6 x 5 x 4) ways = 120 ways.
Then, by fundamental principle, total no of ways on which 3 person can be seated in 6 seats in a row is (6 x 5 x 4) ways = 120 ways.
Question 2: Three men have 3 coats, 4 waist coats and 5 caps. In how many ways can they wear them?
Solution:
Solution:
Three men can wear 3 coats in $^3P_3$ ways. Three men can wear 4 waist coats in $^4P_3$ ways and three men can wear 5 caps in $^5P_3$ ways.
Hence, by fundamental principle, the required number of ways = $^3P_3$ x $^4P_3$ x $^5P_3$ = 8640 ways.
Hence, by fundamental principle, the required number of ways = $^3P_3$ x $^4P_3$ x $^5P_3$ = 8640 ways.
Question 3: In how many different ways can the letters of the word "FLOWERS" be arranged?
Solution:
Solution:
All the letters are distinct that are 7 in number. The required number of permutations is $^7P_7$ = 5040 ways.
Question 4: In how many different ways can 2 boys and 4 girls be arranged on a row, such that all the boys stand together and all the girls stand together?
Solution:
Solution:
2 boys are always together and can be arranged among themselves $^2P_2$ in ways =2! ways.
The 4 girls are always together and can be arranged among themselves in $^4P_4$ ways = 4! ways.
The boys and girls can be arranged among themselves in $^2P_2$ ways = 2! ways.
So, total possible arrangements = 2! x 4! x 2! ways = 96 ways.
The 4 girls are always together and can be arranged among themselves in $^4P_4$ ways = 4! ways.
The boys and girls can be arranged among themselves in $^2P_2$ ways = 2! ways.
So, total possible arrangements = 2! x 4! x 2! ways = 96 ways.