Learn Permutation and Combination Examples Comprehensive

Permutation and Combination Examples and Questions is an important topic to be covered in competitive exams preparation be it for IBPS Clerk, PO, RRB, SO, or SBI Clerk or PO Exams. The number of questions and the complexity of questions vary from exam to exam every year.

Permutation and Combination Examples and Questions are complex in nature and require an understanding of Squares and Square Roots, Cubes and Cube Roots, Math Tables, and Probability. Permutation and Combination Examples and Practice Questions based on Permutation and Combination Examples are given in this article.

Nowadays, questions in exams are mixed with multiple concepts and it requires practice and a deep understanding of basic concepts given in Permutation and Combination Examples along with quickly identifying numbers.

What is Permutation?

A Permutation is a way to find all the possible numbers of arrangements in a definite number of order or all at one time.

When all Objects are Distinct

The number of possible ways in which we can arrange all distinct objects at a time. The number of possibilities of n different objects taken r at a time, where 0 < r < n + 1 and the objects do not repeat is n ( n – 1) ( n – 2)…( n – r + 1), which is denoted by ⁿP_r.

ⁿP_r expression is difficult so it’s being replaced with notation n! known as factorial n or n factorial, which also reduces the size of expression. The notation n! represents the product of first n natural numbers.

Factorial of 10 Numbers

1! = 1
2! = 2 x 1 = 2
3! = 3 x 2 x 1 = 6
4! = 4 x 3 x 2 x 1 = 24
5! = 5 x 4 x 3 x 2 x 1 = 120
6! = 6 x 5 x 4 x 3 x 2 x 1 = 720
7! = 7 x 6 x 5 x 4 x 3 x 2 x 1 = 5,040
8! = 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 40,320
9! = 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 3,62,880
10! = 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 36,28,800

Candidates should memorise these factorials for getting faster solution during exams.

When some Objects are Distinct

The number of permutations of n objects, where p objects are of the same kind and the rest is all different = n! / p!. Also, in a possible scenario where we have multiple repeating objects such as p1, p2 then the permutation of n objects will be = n! / (p1! x p2!).

What is Combination?

The combination is a way of selecting items from a collection, such that the order of selection does not matter. The formula for finding the number of combinations of n different objects taken r at a time, denoted by ⁿC_r.

It can also be calculated as: ⁿP_r = ⁿC_r x r!

Useful Combinations

1. ⁿC_r = n! / (r! x (n – r)!)
2. ⁿC₁ = n
3. ⁿC_n-1 = n
4. ⁿC_n = 1
5. ⁿC₀ = 1

Permutation and Combination Examples

Below are the Permutation and Combination Examples:

Example 1: Find the value of 7! – 5!?

Solution 1: Substituting the value of 7! is 5040 and 5! is 120.
=> 5040 – 120 = 4920.

Example 2: Find value of 7! / 5!?

Solution 2: We can simplify 7! as 7 x 6 x 5! upon dividing 5! with 5! it’ll result in 1.
=> (7 x 6 x 5!) / 5!
=> 7 x 6 = 42

Example 3: Find the number of permutations for the letter ALLAHABAD?

Solution 3: Word consists of 9 letters and there are a few common letters like ‘A’ and ‘L’. The count of ‘A’ is 4 and ‘L’ is 2.
Now finding permutation for word => 9! / (4! x 2!)
=> 9 x 8 x 7 x 6 x 5 x 4! / (4! x 2!)
=> 9 x 8 x 7 x 6 x 5 / 2
=> 15,120 / 2
=> 7,560

Example 4: The number of different 8-letter arrangements that can be made from the letters of the word DAUGHTER so that (i) all vowels occur together and (ii) all vowels do not occur together.

Solution 4: There are a total of 8 letters in the word DAUGHTER and it has 3 Vowels such as ‘A’, ‘E’, and ‘U’. As we’re considering all vowels as 1 so the count for this word will become 6. In a word, 5 letters are consonant and 1 for all remaining vowels as they’ll be placed together.

Solution for (i): 6! x 3! => 720 x 6 => 4320
Solution for (ii): 8! – 6! x 3! => 40320 – 720 x 6 => 40320 – 4320 = 36,000

Example 5: A committee of 3 persons is to be constituted from a group of 2 men and 3 women. (I) In how many ways can this be done? (II) How many of these committees would consist of 1 man and 2 women?

Solution 5: For (I): ⁵C₃ = 5! / (3! x (5-3)!)
=> 5! / (3! x 2!)
=> 5 x 4 / 2
=> 20 / 2 = 10.
For (II): 1 man can be selected from 2 men in ²C₁ ways = 2
2 women can be selected from 3 women in ³C₂ ways = 3
Therefore, total ways to select committee member is = ²C₁ x ³C₂ = 3 x 2 = 6.

Practice Questions on Permutation and Combination Examples

Below are the Practice Questions based on Permutation and Combination Examples:

1. Find the value of 7! – 6!?
2. Find the value of 3! x 5! – 4!?
3. Find the value of 12! / (10! x 2!)?
4. Find the value of 15! / (9! x 5!)?
5. Find the number of permutations for the letter RAJAJI?
6. Find the number of permutations for the letter OPPORTUNITY?
7. Find the number of arrangements of the letters of the word INDEPENDENCE. In how many of these arrangements, (i) do the words start with P (ii) do all the vowels always occur together (iii) do the vowels never occur together (iv) do the words begin with I and end in P?
8. In how many ways can 4 red, 3 yellow, and 2 green discs be arranged in a row if the discs of the same color are indistinguishable?
9. What is the number of ways of choosing 4 cards from a pack of 52 playing cards? In how many of these (i) four cards are of the same suit, (ii) four cards belong to four different suits, (iii) are face cards, (iv) two are red cards and two are black cards, (v) cards are of the same color?
10. A group consists of 4 girls and 7 boys. In how many ways can a team of 5 members be selected if the team has (i) no girl? (ii) at least one boy and one girl? (iii) at least 3 girls?

Final Words

Practicing the above Permutation and Combination Examples and Questions is not the end of your practice, but it’s the start of a new journey to apply logic on Permutation and Combination Examples and Questions with multiple approaches in a right and faster way.

For cracking competitive exams one must practice Permutation and Combination Examples and Questions without a calculator is a must. At last, during the exam, if a solution for the Permutation and Combination Questions cannot be found easily then mark that question to revisit and move ahead instead of wasting time and energy.