Permutations, combinations and probability: Class 11 Mathematics Notes, Short Questions and Numericals

The counting principle, factorials, permutations and combinations with their formulas, arrangements with repeated objects, sample space and events, the definition of probability, and the addition rule for events.

Key points

The fundamental counting principle says that if one task can be done in m ways and a second task in n ways, then the two tasks together can be done in m × n ways. This is the multiplication rule. If the tasks are alternatives, so that only one of them is done, the number of ways is m + n. This is the addition rule.

The factorial of a natural number n is n! = n(n − 1)(n − 2) … 3 × 2 × 1. By definition 0! = 1. For example 5! = 120. Factorials grow very fast, and they are used in all the formulas of this chapter.

A permutation is an arrangement of objects in a definite order. The number of permutations of n different objects taken r at a time is ⁿPᵣ = n!/(n − r)!. When all n objects are used, ⁿPₙ = n!. For example, the number of ways to arrange 3 of 5 books on a shelf is ⁵P₃ = 5 × 4 × 3 = 60.

Special cases of permutations:

Objects with repeats: if n objects include p of one kind, q of another kind and r of a third kind, the number of arrangements is n!/(p! q! r!).
Circular arrangements of n different objects: (n − 1)!, since rotations look the same.
Conditions such as a fixed first or last place: fill that place first, then arrange the rest.

A combination is a selection of objects in which the order does not matter. The number of combinations of n different objects taken r at a time is ⁿCᵣ = n!/[r!(n − r)!]. Each combination of r objects can be arranged in r! ways, so ⁿPᵣ = r! × ⁿCᵣ. Choosing a committee is a combination, while choosing a president and a secretary is a permutation.

Useful facts: ⁿCᵣ = ⁿCₙ₋ᵣ, ⁿC₀ = ⁿCₙ = 1 and ⁿC₁ = n. Also ⁿCᵣ + ⁿCᵣ₋₁ = ⁿ⁺¹Cᵣ. When a selection has two groups, such as men and women, multiply the number of ways of choosing from each group. For at least or at most conditions, add the cases, or subtract the unwanted cases from the total.

Probability measures how likely an event is. An experiment is a process with an outcome that is not certain. The sample space S is the set of all possible outcomes, and an event E is a subset of S. Tossing a coin has S = {H, T}. Rolling a die has S = {1, 2, 3, 4, 5, 6}. Rolling two dice has 36 equally likely outcomes.

When the outcomes are equally likely, the probability of an event E is P(E) = n(E)/n(S). Always 0 ≤ P(E) ≤ 1. An impossible event has probability 0 and a sure event has probability 1. The complement of E, written E′, has P(E′) = 1 − P(E). The probabilities of all the outcomes of an experiment add to 1.

Two events are mutually exclusive if they cannot occur together, so E ∩ F = ∅. For such events P(E ∪ F) = P(E) + P(F). For any two events the addition rule is P(E ∪ F) = P(E) + P(F) − P(E ∩ F). For example, the chance of drawing a king or a heart from a pack of 52 cards must subtract the king of hearts, which is counted twice.

Two events are independent if the occurrence of one does not change the probability of the other. For independent events P(E ∩ F) = P(E) × P(F). A coin toss and a die roll are independent. The conditional probability of E given F is P(E | F) = P(E ∩ F)/P(F), when P(F) ≠ 0.

Study tip: Before any counting problem, ask whether the order matters. For probability, list the sample space size first, then count the favourable outcomes carefully, using a table for two dice. For at least questions, try the complement first, since 1 − P(E′) is often shorter.

Definitions and flashcards

Say the meaning first, then open the card to check.

Factorial

The product n! of all natural numbers from 1 to n, with 0! = 1.

Permutation

An arrangement of objects in a definite order.

Combination

A selection of objects where the order does not matter.

Sample space

The set of all possible outcomes of an experiment.

Event

A subset of the sample space.

Mutually exclusive events

Events that cannot happen at the same time.

Independent events

Events where one does not affect the probability of the other.

Complement of an event

The event that E does not occur, written E′.

Formula sheet

Name	Formula	What it means
Permutations	ⁿPᵣ = n!/(n − r)!	Order matters.
Combinations	ⁿCᵣ = n!/[r!(n − r)!]	Order does not matter.
Link	ⁿPᵣ = r! × ⁿCᵣ	Arrange after choosing.
Repeated objects	n!/(p! q! r!)	p, q, r are the numbers of identical objects.
Probability	P(E) = n(E)/n(S)	Equally likely outcomes.
Complement	P(E′) = 1 − P(E)	Probability of not E.
Addition rule	P(E ∪ F) = P(E) + P(F) − P(E ∩ F)	For any two events.
Independent events	P(E ∩ F) = P(E) × P(F)	Both events occur.

Short questions with model answers

These are the kind of questions asked for 2, 3 and 5 marks. Try each one before opening the answer.

State the difference between a permutation and a combination. (2 marks)

A permutation is an arrangement in which the order of the objects matters, so ABC and CBA are different. A combination is a selection in which the order does not matter, so ABC and CBA are the same selection. For the same n and r, the number of permutations is r! times the number of combinations.

Evaluate ⁷C₂ and ⁷P₂. (2 marks)

⁷P₂ = 7 × 6 = 42. ⁷C₂ = 42/2! = 21.

How many different arrangements are there of the letters of the word BALLOON? (3 marks)

BALLOON has 7 letters. L appears 2 times and O appears 2 times, and B, A, N appear once. The number of arrangements is 7!/(2! × 2!) = 5040/4 = 1260.

State the addition rule of probability and explain when it simplifies. (3 marks)

For any two events E and F, P(E ∪ F) = P(E) + P(F) − P(E ∩ F). The overlap is subtracted because it was counted twice. If E and F are mutually exclusive, then E ∩ F is empty and P(E ∩ F) = 0, so P(E ∪ F) = P(E) + P(F).

Prove that ⁿPᵣ = n!/(n − r)! using the counting principle. (5 marks)

To arrange r objects out of n in a row, fill r places in turn. The first place can be filled in n ways. The second can be filled in n − 1 ways, since one object is used. The third in n − 2 ways, and so on, until the rth place, which can be filled in n − r + 1 ways. By the multiplication rule, ⁿPᵣ = n(n − 1)(n − 2) … (n − r + 1). Multiply and divide by (n − r)!, and the numerator becomes n!. So ⁿPᵣ = n!/(n − r)!.

Two fair dice are rolled. Find the probability that the sum is 9 and the probability that the sum is at least 10. (5 marks)

The sample space has 36 equally likely outcomes. A sum of 9 occurs for (3, 6), (4, 5), (5, 4) and (6, 3), which is 4 outcomes, so the probability is 4/36 = 1/9. A sum of 10 has 3 outcomes, 11 has 2 outcomes and 12 has 1 outcome, giving 6 outcomes in all. So the probability of at least 10 is 6/36 = 1/6.

Numericals with worked steps

Find the number of arrangements of the letters of the word LEVEL.

LEVEL has 5 letters: L twice, E twice and V once.
Number of arrangements = 5!/(2! × 2!).
= 120/4 = 30.

Answer: 30 arrangements.

A committee of 4 is chosen from 6 men and 4 women. Find the number of committees with exactly 2 women.

Choose 2 women from 4: ⁴C₂ = 6.
Choose 2 men from 6: ⁶C₂ = (6 × 5)/2 = 15.
Multiply the two independent choices: 6 × 15 = 90.

Answer: 90 committees.

How many 5 digit even numbers can be formed from the digits 1, 2, 3, 4, 5 without repeating a digit?

The number is even if its last digit is 2 or 4, which gives 2 choices.
The remaining 4 places use the other 4 digits in any order: 4! = 24 ways.
Total = 2 × 24 = 48.

Answer: 48 numbers.

A card is drawn from a pack of 52. Find the probability that it is a king or a heart.

P(king) = 4/52 and P(heart) = 13/52.
The king of hearts is in both, so P(king and heart) = 1/52.
P(king or heart) = 4/52 + 13/52 − 1/52 = 16/52.
Simplify: 16/52 = 4/13.

Answer: 4/13.

Common mistakes

Using a permutation when the problem asks for a selection, or a combination when the order matters.
Forgetting to divide by the factorials of repeated letters when arranging words such as LEVEL.
Adding P(E) and P(F) for events that are not mutually exclusive, without subtracting the overlap.
Writing 0! = 0. The correct value is 0! = 1.
Multiplying when the problem needs adding of cases, or adding when the tasks happen together.

Practise this chapter

Practice questions and timed test Printable worksheet