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Algebra - Permutations and Combinations

Grade 11ICSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Fundamental Principle of Counting: The Multiplication Principle states that if one event can occur in mm ways and a second event can occur in nn ways, then the total number of ways both events can occur in sequence is m×nm \times n. The Addition Principle states that if two events are mutually exclusive, they can occur in m+nm + n ways. Visually, think of the Multiplication Principle as a tree diagram where each choice at the first level branches out into multiple choices at the second level.

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Factorial Notation: The expression n!n! (read as nn factorial) represents the product of the first nn natural numbers, where n!=n×(n−1)×(n−2)×⋯×1n! = n \times (n-1) \times (n-2) \times \dots \times 1. By convention, 0!=10! = 1. This concept can be visualized as the total number of ways to arrange nn items in a straight line.

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Permutations (Arrangements): Permutations refer to the different ways of arranging a set of objects where the order of arrangement is critical. For example, the arrangement ABAB is different from BABA. This is visually similar to placing objects into specific, numbered slots where the identity of the slot matters.

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Combinations (Selections): Combinations refer to the selection of items from a larger set where the order of selection does not matter. For example, selecting a committee of 2 people from {A, B, C} results in {A, B}, {A, C}, and {B, C} (where {A, B} is the same as {B, A}). Visually, this is like putting selected items into a single un-ordered bag.

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Circular Permutations: When objects are arranged in a circle, the number of distinct arrangements is (n−1)!(n-1)!. This is because rotating the circle does not change the relative positions of the objects. Visually, to find the number of arrangements, we 'fix' one object at the top of the circle to provide a reference point, and then arrange the remaining (n−1)(n-1) objects linearly.

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Permutations with Identical Objects: If there are nn objects where pp objects are of one kind, qq objects are of another kind, and so on, the number of arrangements is given by n!p!q!…\frac{n!}{p!q!\dots}. Visually, this accounts for the fact that swapping two identical objects doesn't create a 'new' looking arrangement, so we divide by the number of ways those identical items can be permuted among themselves.

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Restricted Permutations: In problems where certain items must be together, we treat that group of items as a single 'block' or entity. After arranging the blocks, we then multiply by the number of ways to arrange the items inside that specific block. Visually, imagine tying the items together with a string to form a single unit before placing them into the sequence.

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Relationship between nPr^nP_r and nCr^nC_r: The number of permutations of nn things taken rr at a time is equal to the number of combinations of nn things taken rr at a time multiplied by the ways to arrange those rr things. This is expressed as nPr=nCr×r!^nP_r = ^nC_r \times r!.

📐Formulae

n!=n(n−1)(n−2)…3⋅2⋅1n! = n(n-1)(n-2)\dots 3 \cdot 2 \cdot 1

nPr=n!(n−r)!^nP_r = \frac{n!}{(n-r)!}

nCr=n!r!(n−r)!^nC_r = \frac{n!}{r!(n-r)!}

nCr=nCn−r^nC_r = ^nC_{n-r}

nCr+nCr−1=n+1Cr^nC_r + ^nC_{r-1} = ^{n+1}C_r

Number of circular permutations=(n−1)!\text{Number of circular permutations} = (n-1)!

Permutations of n objects with repetitions=n!p!q!r!\text{Permutations of } n \text{ objects with repetitions} = \frac{n!}{p!q!r!}

💡Examples

Problem 1:

How many words can be formed using all the letters of the word 'EQUATION' so that the vowels always occur together?

Solution:

Step 1: Identify the vowels and consonants in 'EQUATION'. Vowels are {E, U, A, I, O} (5 letters) and consonants are {Q, T, N} (3 letters). Step 2: Since vowels must be together, treat the 5 vowels as a single block. Now we have the block (EUAIO) plus the 3 consonants, making a total of 1+3=41 + 3 = 4 units to arrange. Step 3: The 4 units can be arranged in 4!4! ways. Step 4: Within the vowel block, the 5 vowels can be arranged among themselves in 5!5! ways. Step 5: Total arrangements = 4!×5!=24×120=28804! \times 5! = 24 \times 120 = 2880.

Explanation:

We use the 'string method' or 'block method' to handle the restriction. First, we arrange the external units, then account for the internal arrangements of the constrained group.

Problem 2:

A committee of 5 is to be formed from 6 gentlemen and 4 ladies. In how many ways can this be done if the committee must contain at least 3 ladies?

Solution:

Step 1: Identify the possible cases for 'at least 3 ladies' in a committee of 5. Case 1: 3 ladies and 2 gentlemen. Case 2: 4 ladies and 1 gentleman. Step 2: Calculate combinations for Case 1: 4C3×6C2=4×15=60^4C_3 \times ^6C_2 = 4 \times 15 = 60. Step 3: Calculate combinations for Case 2: 4C4×6C1=1×6=6^4C_4 \times ^6C_1 = 1 \times 6 = 6. Step 4: Add the results of the mutually exclusive cases: 60+6=6660 + 6 = 66.

Explanation:

Since the order of selection for the committee doesn't matter, we use combinations. 'At least' implies we must sum the results of all possible valid scenarios.