This MCQ module is based on: 6.1 Introduction
TOPIC 18 OF 45
6.1 Introduction
🎓 Class 11
Mathematics
CBSE
Theory
Ch 6 — Permutations and Combinations
⏱ ~16 min
🌐 Language: [gtranslate]
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This mathematics assessment will be based on: 6.1 Introduction
Targeting Class 11 level in Combinatorics, with Advanced difficulty.
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6.1 Introduction
Suppose you have a suitcase with a number lock. The number lock has 4 wheels each labelled with 10 digits from 0 to 9. The lock can be opened if 4 specific digits are arranged in a particular sequence with no repetition. Some of you may have forgotten this specific sequence of digits. You remember only the first digit which is 7. In order to open the lock, how many sequences of 3-digits you may have to check with? To answer this question, you may, immediately, start listing all possible arrangements of 9 remaining digits taken 3 at a time. But, this method will be tedious, because the number of possible sequences may be large. Here, in this Chapter, we shall learn some basic counting techniques which will enable us to answer this question without actually listing 3-digit arrangements.
In fact, these techniques will be useful in determining the number of different ways of arranging and selecting objects without actually listing them. As a first step, we shall examine a principle which is most fundamental to the learning of these techniques.
Jacob Bernoulli
(1654 – 1705)
Swiss mathematician. Ars Conjectandi (1713) gave the first full treatment of permutations and combinations.
6.2 Fundamental Principle of Counting
Let us consider the following problem. Mohan has 3 pants and 2 shirts. How many different pairs of a pant and a shirt, can he dress up with? There are 3 ways in which a pant can be chosen, because there are 3 pants available. Similarly, a shirt can be chosen in 2 ways. For every choice of a pant, there are 2 choices of a shirt. Therefore, there are \(3 \times 2 = 6\) pairs of a pant and a shirt.
Let us name the three pants as \(P_1, P_2, P_3\) and the two shirts as \(S_1, S_2\). Then, these six possibilities can be illustrated in the Fig. 6.1.
Fig 6.1 — Tree diagram for Mohan's 3 pants × 2 shirts = 6 outfit combinations.
Now suppose Sabnam has 2 school bags, 3 tiffin boxes and 2 water bottles. In how many ways can she carry these items (choosing one each)? There are 2 ways of choosing a school bag. For each school bag, there are 3 ways of choosing a tiffin box. Thus, there are \(2 \times 3 = 6\) pairs of school bag and tiffin box. Then for each such pair, there are 2 water bottles. So the number of total ways is \(2 \times 3 \times 2 = 12\).
Fig 6.2 — 2 bags × 3 tiffins × 2 water bottles = 12 possibilities.
In fact, the problem of the above types are solved by applying the following principle known as the fundamental principle of counting, or, simply, the multiplication principle?, which states that:
Fundamental Principle of Counting
"If an event can occur in \(m\) different ways, following which another event can occur in \(n\) different ways, then the total number of occurrences of the events in the given order is \(m \times n\)."The above principle can be generalised for any finite number of events. For example, for 3 events, the principle is as follows: "If an event can occur in \(m\) different ways, following which another event can occur in \(n\) different ways, following which a third event can occur in \(p\) different ways, then the total number of occurrences to 'the events in the given order' is \(m \times n \times p\)."
Worked Examples
Example 1. Find the number of 4-letter words, with or without meaning, which can be formed out of the letters of the word ROSE, where the repetition of the letters is not allowed.
Four places (boxes) to fill with 4 distinct letters. First box: 4 choices. After placing one letter, second box: 3 choices. Then 2, then 1. By the multiplication principle: \(4 \times 3 \times 2 \times 1 = 24\) words.
Example 2. Given 4 flags of different colours, how many different signals can be generated, if a signal requires the use of 2 flags one below the other?
Upper flag: 4 choices. For each, lower flag (a different colour): 3 choices. Total signals = \(4 \times 3 = 12\).
Example 3. How many 2 digit even numbers can be formed from the digits 1, 2, 3, 4, 5 if the digits can be repeated?
Units place must be even — choices: 2 or 4, i.e. 2 ways. Tens place: any of 5 digits (repetition allowed). Total = \(5 \times 2 = 10\) numbers.
Example 4. Find 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, (ii) all vowels do not occur together.
DAUGHTER has vowels {A, U, E} and consonants {D, G, H, T, R}. (i) Treat {AUE} as a single block: then 6 units arranged in \(6!\) ways, and the vowels arranged internally in \(3!\) ways. Total = \(6! \times 3! = 720 \times 6 = 4320\). (ii) Total unrestricted arrangements = \(8! = 40320\); so "vowels not all together" = \(40320 - 4320 = 36000\).
Exercise 6.1 — Selected Questions
Q1. How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that (i) repetition of the digits is allowed? (ii) repetition is not allowed?
(i) \(5 \times 5 \times 5 = 125\). (ii) \(5 \times 4 \times 3 = 60\).
Q2. How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the digits can be repeated?
Units place (even): 2, 4, 6 → 3 ways. Tens and hundreds: 6 ways each. Total = \(6 \times 6 \times 3 = 108\).
Q3. How many 4-letter code can be formed using the first 10 letters of the English alphabet, if no letter can be repeated?
\(10 \times 9 \times 8 \times 7 = 5040\) codes.
Q4. How many 5-digit telephone numbers can be constructed using the digits 0 to 9 if each number starts with 67 and no digit appears more than once?
First two fixed (6 and 7). Remaining 3 places filled from remaining 8 digits: \(8 \times 7 \times 6 = 336\).
Q5. A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes are there?
\(2 \times 2 \times 2 = 8\) outcomes.
Q6. Given 5 flags of different colours, how many different signals can be generated if each signal requires the use of 2 flags, one below the other?
\(5 \times 4 = 20\) signals.
Activity 6.1 — Design a Menu Tree
Predict: A café offers 3 starters, 4 main courses, and 2 desserts. Without drawing, how many different 3-course meals are possible?
- Write the three "events": Starter, Main, Dessert — with counts 3, 4, 2.
- Draw a small tree: 3 starter branches → each branches into 4 mains → each branches into 2 desserts.
- Count the leaves at the bottom of the tree.
- Verify by multiplication principle: \(3 \times 4 \times 2 = 24\).
- Now change the problem: if the café adds 2 drinks, how many different 4-course orderings? Compute \(3 \times 4 \times 2 \times 2 = 48\).
Insight: Each new independent choice multiplies the count. Adding a single 2-option choice doubled 24 to 48 — the factor of 2 came straight from the multiplication principle.
Competency-Based Questions — Counting Scenarios
A school cybersecurity club is designing 4-character passcodes for its lab computers. The first character must be a capital letter (A–Z) and the remaining three must be digits (0–9).
Q1. How many distinct passcodes are possible if digits can repeat?
L3 Apply\(26 \times 10 \times 10 \times 10 = 26000\).
Q2. If the three digits must all be distinct, how many passcodes are possible?
L4 Analyse\(26 \times 10 \times 9 \times 8 = 18720\).
Q3. The club wants at least 1 lakh (100000) distinct passcodes. Is a 4-character passcode enough (with repetition)? Justify.
L5 EvaluateMaximum with repetition = \(26 \times 10^3 = 26000\). This is below 1 lakh — so 4 characters are not enough. A 5-character passcode (still letter + 4 digits) gives \(26 \times 10^4 = 260000 \geq 100000\) ✓.
Q4. Design a rule that yields exactly 60000 passcodes. Justify using the counting principle.
L6 CreateOne valid design: letter ∈ {vowels A, E, I, O, U} (5 choices), then 4 digits with repetition → \(5 \times 10^4 = 50000\) (too few). Try \(6 \times 10000 = 60000\): first character from 6 vowels+Y set, remaining 4 digits free. Total = 60000.
Assertion–Reason Questions
Assertion (A): The number of 2-digit numbers formed using digits 1–5 with repetition allowed is 25.
Reason (R): By the multiplication principle, two independent choices each with 5 options give \(5 \times 5\) outcomes.
Reason (R): By the multiplication principle, two independent choices each with 5 options give \(5 \times 5\) outcomes.
Answer: A. Both true; R directly supplies the justification for A.
Assertion (A): Tree diagrams and the multiplication principle always give the same count.
Reason (R): A tree diagram enumerates every outcome of a sequence of independent choices, so counting leaves is equivalent to multiplying branch counts.
Reason (R): A tree diagram enumerates every outcome of a sequence of independent choices, so counting leaves is equivalent to multiplying branch counts.
Answer: A. Counting leaves of a uniform branching tree equals the product of branch counts — the two methods are equivalent.
Assertion (A): The number of 3-digit numbers with all digits distinct that can be formed from 0–9 is \(10 \times 9 \times 8 = 720\).
Reason (R): A 3-digit number cannot have 0 in the hundreds place.
Reason (R): A 3-digit number cannot have 0 in the hundreds place.
Answer: D. A is false because the hundreds place has only 9 valid choices (not 10); correct count is \(9 \times 9 \times 8 = 648\). R is a true supporting fact.
Frequently Asked Questions — Permutations and Combinations
What is Part 1 — Permutations and Combinations: Fundamental Principle of Counting | Class 11 Maths | MyAiSchool in NCERT Class 11 Mathematics?
Part 1 — Permutations and Combinations: Fundamental Principle of Counting | Class 11 Maths | MyAiSchool is a key concept covered in NCERT Class 11 Mathematics, Chapter 6: Permutations and Combinations. This lesson builds the student's foundation in the chapter by explaining the core ideas with worked examples, definitions, and step-by-step methods aligned to the CBSE curriculum.
How do I solve problems on Part 1 — Permutations and Combinations: Fundamental Principle of Counting | Class 11 Maths | MyAiSchool step by step?
To solve problems on Part 1 — Permutations and Combinations: Fundamental Principle of Counting | Class 11 Maths | MyAiSchool, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 11 exercises gradually increase in difficulty — start with solved NCERT examples before attempting exercise questions, and always verify your answer by substitution or diagram.
What are the most important formulas for Chapter 6: Permutations and Combinations?
The essential formulas of Chapter 6 (Permutations and Combinations) are listed in the chapter summary and highlighted throughout the lesson in formula boxes. Memorise them and practise at least 2–3 problems per formula. CBSE board exams frequently test direct application as well as combined use of multiple formulas from this chapter.
Is Part 1 — Permutations and Combinations: Fundamental Principle of Counting | Class 11 Maths | MyAiSchool important for the Class 11 board exam?
Part 1 — Permutations and Combinations: Fundamental Principle of Counting | Class 11 Maths | MyAiSchool is part of the NCERT Class 11 Mathematics syllabus and appears in CBSE board exams. Questions typically include short-answer, long-answer, and competency-based items. Review the NCERT examples, exercise questions, and previous-year board problems on this topic to prepare confidently.
What mistakes should students avoid in Part 1 — Permutations and Combinations: Fundamental Principle of Counting | Class 11 Maths | MyAiSchool?
Common mistakes in Part 1 — Permutations and Combinations: Fundamental Principle of Counting | Class 11 Maths | MyAiSchool include skipping steps, misapplying formulas, sign errors, and losing track of units. Write each step clearly, double-check algebraic manipulations, and re-read the question after solving to verify that your answer matches what was asked.
Where can I find more NCERT practice questions on Part 1 — Permutations and Combinations: Fundamental Principle of Counting | Class 11 Maths | MyAiSchool?
End-of-chapter NCERT exercises for Part 1 — Permutations and Combinations: Fundamental Principle of Counting | Class 11 Maths | MyAiSchool cover all difficulty levels tested in CBSE exams. After completing them, try the examples again without looking at the solutions, attempt the NCERT Exemplar questions for Chapter 6, and solve at least one previous-year board paper to consolidate your understanding.
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