Permutation and Combination: Definition, Formula, Solved Questions

Permutation and Combination: Permutation and Combination are one of the most important concepts in Mathematics. It is used in the practical applications in science, engineering and research, and many other fields. The difference between permutation and combination can be defined as; when a set of data is selected from a certain group, it is known as permutation; while the order in which the data is arranged is known as combination. Thus, we can say that permutation is an ordered combination.

Grouping of data and its interpretation is possible with the help of permutation and combination. Mathematics students should be well versed in the concepts because it finds applications in all advanced research and data analysis. This article provides insight on the permutation and combination questions, solutions, permutation and combination formulas and more.

Permutation and Combination Definition

Here we have given the mathematical definition of permutation and combination:

What is Permutation?

A permutation is an arrangement in a definite order of several objects taken, some or all at a time, with permutations, every tiny detail matters. It means the order in which elements are arranged is significant.

There are two types of permutations:

• Repetition is Allowed: For the number lock example provided above, it could be “2-2-2”.
• No Repetition Allowed: For example, the first three people in a race. You can’t be first and second at the same time.
  

What is Combination?

The combination is a way of selecting elements from a set so that the order of selection doesn’t matter. With the combination, only choosing elements matters. It means the order in which elements are chosen is not essential.

There are two types of combinations:

• Repetition is Allowed: For example, coins in your pocket (2, 5, 5, 10, 10)
• No Repetition Allowed: For example, lottery numbers (2, 14, 18, 25, 30, 38)

Permutation and Combination Formula

There are many formulas that are used to solve permutation and combination problems. We have provided the complete permutation formula list here:

• When repetition is not allowed: P is a permutation or arrangement of r things from a set of n things without replacement.
• When repetition is allowed: P is a permutation or arrangement of r things from a set of n things when duplication is allowed.

Derivation of Permutation Formula: Let us assume that there are r boxes, and each of them can hold one thing. There will be as many permutations as there are ways of filling in r vacant boxes by n objects.

• No. of ways the first box can be filled: n

• No. of ways the second box can be filled: (n – 1)

• No. of ways the third box can be filled: (n – 2)

• No. of ways the fourth box can be filled: (n – 3)

• No. of ways r^th box can be filled: [n – (r – 1)]

• The number of permutations of n different objects taken r at a time, where 0 < r ≤ n and the objects do not repeat is: n(n – 1)(n – 2)(n – 3) . . . (n – r + 1)

We have provided the complete combination formula list here:

• When repetition is not allowed: C is a combination of n distinct things taking r at a time (order is not important).
• When repetition is allowed: C is a combination of n distinct things taking r at a time (order is not important) with repetition. We define C as:
  

Derivation of Combination Formula:

Let us assume that there are r boxes, and each of them can hold one thing.

• No. of ways to select the first object from n distinct objects: n
• No. of ways to select the second object from (n-1) distinct objects: (n-1)
• No. of ways to select the third object from (n-2) distinct objects: (n-2)
• No. of ways to select r^th object from [n-(r-1)] distinct objects: [n-(r-1)]

Completing the selection of r things from the original set of n things creates an ordered subset of r elements.
∴ The number of ways to make a selection of r elements of the original set of n elements is: n (n – 1) (n – (n-3) . . . (n – (r – 1)) or n (n – 1) (n – 2) … (n – r + 1). 

Let us consider the ordered subset of r elements and all their permutations. The total number of permutations of this subset equals r! because r objects in every combination can be rearranged in r! ways.

Hence, the total number of permutations of n different things taken r at a time is (nC_r×r!). It is nothing but nP_r.

Download – Permutation and Combination Formula PDF

We can summarize the permutation combination formula in the table below:

Difference Between Permutation and Combination

We have provided the permutation and combination differences in the table below:

Check:

Solved Examples of Permutation and Combination

We have provided some permutation and combination examples with detailed solutions. Get Permutation and Combination Class 11 NCERT Solutions for on Embibe.

Q.1: Find the number of permutations and combinations, if n = 15 and r = 3.
Ans: n = 15, r = 3 (Given)
Using the formulas for permutation and combination, we get:
Permutation, P = n!/(n – r)!
= 15!/(15 – 3)!
= 15!/12!
= (15 x 14 x 13 x 12!)/12!
= 15 x 14 x 13
= 2730
Also, Combination, C = n!/(n – r)!r!
= 15!/(15 – 3)!3!
= 15!/12!3!
= (15 x 14 x 13 x 12! )/12!3!
= 15 x 14 x 13/6
= 2730/6
= 455

Q.2: In how many different ways can the letters of the word ‘OPTICAL’ be arranged so that the vowels always come together?
Ans: The word ‘OPTICAL’ has 7 letters. It has the vowels’ O’, ‘I’, and ‘A’ in it and these 3 vowels should always come together. Hence these three vowels can be grouped and considered as a single letter. That is, PTCL(OIA).
Hence we can assume the total letters as 5 and all these letters are different.
Number of ways to arrange these letters
= 5!
= 5×4×3×2×1
= 120
All the 3 vowels (OIA) are different.
Number of ways to arrange these vowels among themselves
= 3!
= 3×2×1
= 6
Hence, the required number of ways:
= 120×6
= 720

Q.3: How many 3-letter words with or without meaning, can be formed out of the letters of the word, ‘LOGARITHMS’ if repetition of letters is not allowed?
Ans: The word ‘LOGARITHMS’ has 10 different letters.
Hence, the number of 3-letter words (with or without meaning) formed by using these letters
= ¹⁰P₃
= 10×9×8
= 720

Q.4: There are 8 men and 10 women and you need to form a committee of 5 men and 6 women. In how many ways can the committee be formed?
Ans: We need to select 5 men from 8 men and 6 women from 10 women.
Number of ways to do this
= ⁸C₅ × ¹⁰C₆
= ⁸C₃ × ¹⁰C₄ [∵ ⁿC_r = ⁿC_(n-r)]

= [(8 x 7 x 6)/(3 x 2 x 1)] x [(10 x 9 x 8 x 7)/(4 x 3 x 2 x 1)]

= 56×210
= 11760

Q.5: A bag contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the bag if at least one black ball is to be included in the draw?
Ans: From 2 white balls, 3 black balls and 4 red balls, 3 balls are to be selected such that at least one black ball should be there.
Hence we have 3 choices as given below.
Choice 1: We can select 3 black balls.
Choice 2: We can select 2 black balls and 1 non-black ball.
Choice 3: We can select 1 black ball and 2 non-black balls.

Number of ways to select 3 black balls
= 3C3
Number of ways to select 2 black balls and 1 non-black ball
= 3C2 × 6C1
Number of ways to select 1 black ball and 2 non-black balls
= 3C1 × 6C2
Total number of ways
= 3C3 + 3C2 × 6C1 + 3C1 × 6C2
= 3C3 + 3C1 × 6C1 + 3C1 × 6C2[∵ ⁿC_r = ⁿC_(n-r)]

= 1 + (3×6) + [3 x (6×5)/(2×1)]

= 1 + 18 + 45
= 64

Q.6: An event manager has ten patterns of chairs and eight patterns of tables. In how many ways can he make a pair of tables and chairs?
Ans: The event manager has 10 patterns of chairs and 8 pattern tables.
A chair can be selected in 10 ways.
A table can be selected in 8 ways.
Hence one chair and one table can be selected in 10 × 8 ways 
= 80 ways

Q.7: In how many ways can three boys can be seated on five chairs?
Ans: There are 3 boys.
The 1st boy can sit in any of the five chairs (5 ways).

Now there are 4 chairs remaining. The second boy can sit in any of the four chairs (4 ways).

Now there are 3 chairs remaining. The third boy can sit in any of the three chairs (3 ways).

Hence, the total number of ways in which 3 boys can be seated on 5 chairs
= 5 × 4 × 3
= 60

Q.8: In how many ways can a team of 5 persons be formed out of a total of 10 persons such that two particular persons should be included in each team?
Answer: Two particular persons should be included in each team. Therefore we have to select the remaining (5 – 2) = 3 persons from (10 – 2) = 8 persons.
Hence, the required number of ways
= ⁸C₃ 
= (8×7×6)/(3×2×1)
= 8×7
= 56

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Permutation and Combination Practice Questions

Here are some practice questions on permutation and combination concepts for you to practice:

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FAQs on Permutation and Combination