Permutations and Combinations

Top

Sub Topics Permutation Combination Repetitions

The terms permutations and combinations both have different meanings. The Permutation is an arrangement of items where its order matters. For example we can prove ML ≠ LM. For instance take any three names of your friends suppose Martin (M), Victor (V), and Bob (B). We can arrange these names in many different orders. If we rearrange them then find out the various possible outcomes: 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 putting the number in the formula: 3! = (3) (2) (1) =6. The formula for calculating permutation is P (n, r) = n! / (n - r)!. Now we come to the Combination; It is a collection of items where its order does not matter in a different arrangement and if order does not matter then ML = LM. Now we will consider the example of Martin, Victor and Bob but in 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) = n! / r! (n - r)!, Now we will use this however the order does not matter C (5, 3) = 5! / 3! (5 - 3)! = 110 / 12 = 9.167. So here we conclude that among combinations and permutations basic difference is of order only. This is all about maths permutations and combinations.

Permutation is an arrangement of different things and symbols in a preferred sequence of particular order. In this, we take different arrangement of a given number of things by taking some or all at a time. This is a brief permutations definition. All permutation consists of letter ‘s’, ‘t’, ‘u’ by taking two at a time will result (st, ts, su, us, tu, ut).

All permutation made with the letter ‘s’, ‘t’, ‘u’ by taking all at a time will give result as (stu, sut, tsu, tus, ust, uts).
Now, we will see the formula for finding the permutation. The formula for calculating the permutations is given as:
For permutation: “P r = n (n - 1) (n - 2)……(n –r + 1) = n! / (n - r) !,
Where, the values of ‘n’ and ‘r’ are the non- negative integers, also the value of ‘r’ is less than 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.

Now we will evaluate permutations definition with help of an example.
Example: Find the number of permutation where value of ‘n’ is 7 and the value of ‘r’ is 4?

Solution: We know that formula for permutation is given by:
Permutation = _np_r = n! / (n – r)!,
Given, the value of n = 7 and value of r = 4.
First we find the factorial of 7.
The factorial of 7 is = 7 * 6 * 5 * 4 * 3 * 2 * 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 * 2 * 1 = 6;
Now divide 5040 by 6;
Permutation = 5040 / 6 = 840.
This is how you can define permutations and solve the problems related with the same.
Combination can be defined as collection of number of objects arranged together but here order does not matters or we can it is un – ordered collection of a unique size. Now we will define combination with help of formulas. Formula to find the combination is given as: Combination = ⁿC_r = n!/ r! (n – r)!, here values of ‘n’ and ‘r’ denote non- negative integers and value of ‘r’ is smaller than equals to value of ‘n’. ‘r’ shows the size of each combination, ‘n’ shows the size of the Set and ‘!’ denotes the factorial operator.

Step 1: Value of ‘r’ and ‘n’ must be known.

Step 2: Then put these values in formula to find value of combination.

Let’s have small introduction of Permutation. Formula to find permutation is given as: Permutation = ⁿp_r = n! / (n – r)!.
Let's understand combination with the help of an example.

Suppose value of ‘n’ is 8 and value of ‘r’ is 5 then find combination and permutation?
Solution: Formulas to find permutation and combination are:
Combination = ⁿC_r = ⁿp_r / r!;
Permutation = ⁿp_r = n! / (n – r)!,
Given, n = 8 and r = 5.
First find the factorial of 8.
Factorial of 8 is = 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 = 40320.
Then find the factorial of (8 – 5);
Factorial of (8 – 5) is = (8 – 5)! = 3!
So factorial of 3 is = 3 * 2 * 1 = 6;
Now divide 40320 by 6;
Permutation = 40320 / 6 = 6720;
Now find factorial of 5.
Factorial of 5 is = 5 * 4 * 3 * 2 * 1 = 120;
Now divide 6720 by 120.
Combination = 6720 / 120 = 56.
This is all about combination definition.
Counting in mathematics depends upon the type of data that is present in the sample Set. This data can have repetitive i.e. two or more values of same data that are found to occur in the set. We use two types of counting methods to define repetitions.
1. Permutation
2. Combination

In Permutation when we discuss repetition we follow a general formula to calculate the number of permutations for any given information.
N^R or N * N * N * N and so on….
Where, 'N' is the number of data that has to be arranged and 'R' is the number of times 'N' will be multiplied to get the permutations of data. When you have 'R' choices to be made out available 'N' things and repetition is allowed, then number of permutations is calculated by multiplying “N” equals to “r” number of times. For example, we have a set of 10 digits (0, 1, 2, 3, 4, 5, 6, 7, 8, 9), such that we have to choose 3 digits out of it to make a 3 digit number. In how many ways it can be done if repetition of digits is allowed?
Solution: In the question we have been given N as 10 and R as 3. The total number of permutations possible for a 3 digit number is: 10³ = 1000.
Next method we follow is Combination. To understand this method with repetition allowed, let us consider an example of it:

Q. Suppose there are 4 flavors of ice-cream A, B, C and D such that we have to select 3 scoops with each of them from the same flavor then how many combinations of scoops are possible?

Solution: Out of 4 ice-cream flavors we can have several possible combinations of 3 scoops:
1. All 3 scoops are of A
2. All 3 scoops are of B
3. All 3 scoops are of C
4. All 3 scoops are of D
So, total number of combinations possible is:
4 * 3 * ⁴C₁ = 48.