Class 11 — Mathematics (NCERT)

Permutations and Combinations

Class 11

✓By the end of this lesson students will be able to understand and apply factorial notation.
✓By the end of this lesson students will be able to define and calculate permutations (nPr) and solve problems involving arrangements.
✓By the end of this lesson students will be able to define and calculate combinations (nCr) and solve problems involving selections.
✓By the end of this lesson students will be able to distinguish between permutations and combinations in various problem scenarios.

Key concepts

Factorial Notation

For a natural number 'n', the factorial n!, is the product of all natural numbers from 1 to n. It is denoted by n!. We define 0! = 1.

n! = n × (n-1) × (n-2) × ... × 3 × 2 × 1

Permutations (nPr)

A permutation is an arrangement of a certain number of objects taken from a given set of objects, where the order of arrangement is important. The number of permutations of 'n' distinct objects taken 'r' at a time is denoted by P(n, r) or nPr.

P(n, r) = n! / (n-r)!

Combinations (nCr)

A combination is a selection of a certain number of objects taken from a given set of objects, where the order of selection is not important. The number of combinations of 'n' distinct objects taken 'r' at a time is denoted by C(n, r) or nCr.

C(n, r) = n! / (r! * (n-r)!)

Key facts to remember

1n! = n × (n-1)! for n > 1.
20! = 1 and 1! = 1.
3P(n, r) = n! / (n-r)! represents the number of permutations of n distinct objects taken r at a time.
4P(n, n) = n! represents the number of permutations of n distinct objects taken all at a time.
5C(n, r) = n! / (r! * (n-r)!) represents the number of combinations of n distinct objects taken r at a time.
6C(n, r) = C(n, n-r) is a useful property for simplifying calculations.
7C(n, 0) = 1 and C(n, n) = 1.
8The relationship between permutations and combinations is P(n, r) = r! × C(n, r).

Worked examples

Example 1

Evaluate the following: (i) 6! (ii) P(8, 3) (iii) C(9, 4)

I(i) To evaluate 6!:

II6! = 6 × 5 × 4 × 3 × 2 × 1

III6! = 720

IV(ii) To evaluate P(8, 3):

VUsing the formula P(n, r) = n! / (n-r)!, we have n=8, r=3.

VIP(8, 3) = 8! / (8-3)! = 8! / 5!

VIIP(8, 3) = (8 × 7 × 6 × 5!) / 5!

VIIIP(8, 3) = 8 × 7 × 6

9P(8, 3) = 336

10(iii) To evaluate C(9, 4):

11Using the formula C(n, r) = n! / (r! * (n-r)!), we have n=9, r=4.

12C(9, 4) = 9! / (4! * (9-4)!) = 9! / (4! * 5!)

13C(9, 4) = (9 × 8 × 7 × 6 × 5!) / ((4 × 3 × 2 × 1) × 5!)

14C(9, 4) = (9 × 8 × 7 × 6) / (4 × 3 × 2 × 1)

15C(9, 4) = (9 × 2 × 7)

16C(9, 4) = 126

Answer

(i) 720 (ii) 336 (iii) 126

Example 2

How many different words can be formed using all the letters of the word 'EQUATION' if each word has 8 letters and no letter is repeated?

IThe word 'EQUATION' has 8 distinct letters.

IIWe need to form words using all 8 letters, which means we are arranging 8 distinct objects taken 8 at a time.

IIIThis is a permutation problem where n = 8 and r = 8.

IVThe number of permutations is P(8, 8) or 8!.

VP(8, 8) = 8!

VI8! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1

VII8! = 40320

Answer

40320 different words can be formed.

When all objects are used for arrangement, the number of permutations is simply n!.

Example 3

A committee of 4 persons is to be formed from a group of 6 men and 5 women. In how many ways can this be done if the committee consists of exactly 3 men and 1 woman?

IWe need to select 3 men from 6 men. This is a combination problem as the order of selection does not matter.

IINumber of ways to select 3 men from 6 men = C(6, 3).

IIIC(6, 3) = 6! / (3! * (6-3)!) = 6! / (3! * 3!)

IVC(6, 3) = (6 × 5 × 4 × 3!) / ((3 × 2 × 1) × 3!)

VC(6, 3) = (6 × 5 × 4) / (3 × 2 × 1)

VIC(6, 3) = 20

VIINext, we need to select 1 woman from 5 women. This is also a combination problem.

VIIINumber of ways to select 1 woman from 5 women = C(5, 1).

9C(5, 1) = 5! / (1! * (5-1)!) = 5! / (1! * 4!)

10C(5, 1) = (5 × 4!) / (1 × 4!)

11C(5, 1) = 5

12By the Fundamental Principle of Counting, the total number of ways to form the committee is the product of the number of ways to select men and women.

13Total ways = C(6, 3) × C(5, 1)

14Total ways = 20 × 5

15Total ways = 100

Answer

The committee can be formed in 100 ways.

When multiple independent selections are made, the total number of ways is found by multiplying the number of ways for each selection.

Common mistakes

✗Confusing permutations with combinations: Students often fail to identify whether the order of arrangement/selection matters in a given problem.
✗Incorrectly calculating factorials, especially forgetting that 0! = 1.
✗Errors in simplifying expressions involving factorials, leading to incorrect numerical answers.
✗Applying the wrong formula (nPr instead of nCr, or vice-versa) based on a misinterpretation of the problem statement.
✗Forgetting to multiply the number of ways for independent selections when forming a group with multiple criteria (e.g., selecting men AND women).

Exam tips

★Carefully read the problem statement to determine if the order of objects is important (permutation) or not (combination). This is the most crucial step.
★Break down complex problems into smaller, manageable parts. For example, if a committee needs men and women, calculate selections for each group separately and then multiply.
★Show all steps clearly, especially when dealing with factorial calculations, to avoid calculation errors and to gain partial marks even if the final answer is incorrect.
★Memorise the key formulas and properties (like C(n, r) = C(n, n-r) and 0! = 1) to save time and improve accuracy during the exam.

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