This MCQ module is based on: 6.4 Combinations
TOPIC 20 OF 45
6.4 Combinations
🎓 Class 11
Mathematics
CBSE
Theory
Ch 6 — Permutations and Combinations
⏱ ~20 min
🌐 Language: [gtranslate]
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This mathematics assessment will be based on: 6.4 Combinations
Targeting Class 11 level in Combinatorics, with Advanced difficulty.
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6.4 Combinations
So far we have arranged objects in different orders — permutations. But in many problems the order does not matter. Suppose a cricket selector must pick 11 players from a squad of 15; the team {Rohit, Virat, Jasprit, …} is the same team no matter the order in which the names are read. This kind of "selection without regard to order" is called a combination?.
Definition — Combination
A combination is a selection of some or all of a number of different objects without regard to order. The number of combinations of \(n\) distinct objects taken \(r\) at a time is denoted by \({}^nC_r\) or \(\binom{n}{r}\).Consider 4 distinct objects a, b, c, d. The combinations taken 3 at a time are only: \(\{a,b,c\}, \{a,b,d\}, \{a,c,d\}, \{b,c,d\}\) — exactly 4 selections. But permutations of 3 objects from 4 number \({}^4P_3 = 24\). The ratio is \(24/4 = 6 = 3!\): each combination gives rise to \(3!\) permutations by reordering its 3 elements. This observation is the heart of the combinations formula.
Derivation of nCr
Every selection of \(r\) distinct objects from \(n\) can be internally reordered in \(r!\) ways to produce \(r!\) permutations. Hence
\[ {}^nP_r = r! \cdot {}^nC_r, \] which rearranges toTheorem 6.3 — Combinations Formula
The number of combinations of \(n\) distinct objects taken \(r\) at a time \((0 \leq r \leq n)\) is
\[ \boxed{\;{}^nC_r = \dfrac{n!}{r!\,(n-r)!}\;} \]
In particular, \({}^nC_0 = 1 = {}^nC_n\) and \({}^nC_1 = n\).Properties of nCr
Theorem 6.4 — Symmetry
\[ {}^nC_r = {}^nC_{n-r} \]Proof. \({}^nC_{n-r} = \dfrac{n!}{(n-r)!\,(n-(n-r))!} = \dfrac{n!}{(n-r)!\,r!} = {}^nC_r\). Intuitively, choosing which \(r\) objects to include is equivalent to choosing which \(n-r\) objects to exclude.
Theorem 6.5 — Pascal's Identity
\[ {}^nC_r + {}^nC_{r-1} = {}^{n+1}C_r \quad (1 \leq r \leq n) \]Proof. \[ {}^nC_r + {}^nC_{r-1} = \dfrac{n!}{r!(n-r)!} + \dfrac{n!}{(r-1)!(n-r+1)!} \] \[ = \dfrac{n!}{(r-1)!(n-r)!}\left[\dfrac{1}{r} + \dfrac{1}{n-r+1}\right] = \dfrac{n!}{(r-1)!(n-r)!} \cdot \dfrac{n+1}{r(n-r+1)} = \dfrac{(n+1)!}{r!(n-r+1)!} = {}^{n+1}C_r. \]
Corollary — Equal Combinations
If \({}^nC_a = {}^nC_b\), then either \(a = b\) or \(a + b = n\).Fig 6.4 — Pascal's triangle: row \(n\) shows \({}^nC_0, {}^nC_1, \ldots, {}^nC_n\); each entry is the sum of the two above (Theorem 6.5).
Interactive: nCr vs nPr
Change \(n\) and \(r\). Notice \({}^nC_r = {}^nP_r / r!\).
nPr = 120
nCr = 20
Worked Examples
Example 13. If \({}^nC_9 = {}^nC_8\), find \({}^nC_{17}\).
By the corollary, either \(9 = 8\) (impossible) or \(9 + 8 = n\), so \(n = 17\). Therefore \({}^nC_{17} = {}^{17}C_{17} = 1\).
Example 14. A committee of 3 persons is to be formed from 7 men and 4 women. In how many ways can this be done when the committee consists of (i) exactly 1 woman, (ii) at least 1 woman, (iii) at most 1 woman?
(i) 1 woman from 4 and 2 men from 7: \({}^4C_1 \cdot {}^7C_2 = 4 \cdot 21 = 84\).
(ii) At least 1 woman = 1W+2M + 2W+1M + 3W+0M = \(4\cdot 21 + 6\cdot 7 + 4\cdot 1 = 84 + 42 + 4 = 130\).
(iii) At most 1 woman = 0W+3M + 1W+2M = \({}^7C_3 + 84 = 35 + 84 = 119\).
(ii) At least 1 woman = 1W+2M + 2W+1M + 3W+0M = \(4\cdot 21 + 6\cdot 7 + 4\cdot 1 = 84 + 42 + 4 = 130\).
(iii) At most 1 woman = 0W+3M + 1W+2M = \({}^7C_3 + 84 = 35 + 84 = 119\).
Example 15. What is the number of ways of choosing 4 cards from a pack of 52 playing cards? In how many of these will (i) all 4 be of the same suit, (ii) all 4 be face cards, (iii) two be red and two be black, (iv) cards be of the same colour?
Total: \({}^{52}C_4 = 270725\).
(i) 4 suits × \({}^{13}C_4\) = \(4 \times 715 = 2860\).
(ii) Face cards = 12; \({}^{12}C_4 = 495\).
(iii) \({}^{26}C_2 \cdot {}^{26}C_2 = 325 \cdot 325 = 105625\).
(iv) Same colour = 2 × \({}^{26}C_4 = 2 \times 14950 = 29900\).
(i) 4 suits × \({}^{13}C_4\) = \(4 \times 715 = 2860\).
(ii) Face cards = 12; \({}^{12}C_4 = 495\).
(iii) \({}^{26}C_2 \cdot {}^{26}C_2 = 325 \cdot 325 = 105625\).
(iv) Same colour = 2 × \({}^{26}C_4 = 2 \times 14950 = 29900\).
Example 16. The English alphabet has 5 vowels and 21 consonants. How many words with 2 different vowels and 2 different consonants can be formed from the alphabet?
Select 2 vowels: \({}^5C_2 = 10\). Select 2 consonants: \({}^{21}C_2 = 210\). Arrange the 4 chosen letters: \(4! = 24\). Total \(= 10 \cdot 210 \cdot 24 = 50400\) words.
Activity 6.3 — Pascal's Triangle Patterns
Predict: What is the sum of the entries in the \(n\)-th row of Pascal's triangle?
- Write rows 0 through 5; compute each row sum: 1, 2, 4, 8, 16, 32.
- Recognise the pattern: row \(n\) sum \(= 2^n\).
- Prove: \(\sum_{r=0}^{n} {}^nC_r = 2^n\) (this counts all subsets of an \(n\)-element set).
- Verify the "hockey stick": \({}^2C_2 + {}^3C_2 + {}^4C_2 + {}^5C_2 = 1+3+6+10 = 20 = {}^6C_3\) ✓.
- Colour odd numbers in the triangle — you'll see the Sierpiński gasket fractal emerge.
Insight: Pascal's triangle compactly encodes every binomial coefficient. Its symmetry mirrors \({}^nC_r = {}^nC_{n-r}\), its row sums count total subsets, and its recursive build rule is Theorem 6.5 in pictures.
Exercise 6.3 — Selected Questions
Q1. If \({}^nC_8 = {}^nC_2\), find \({}^nC_2\).
\(8 + 2 = n \Rightarrow n = 10\). \({}^{10}C_2 = 45\).
Q2. Determine \(n\) if (i) \({}^{2n}C_3 : {}^nC_3 = 12:1\), (ii) \({}^{2n}C_3 : {}^nC_3 = 11:1\).
\(\dfrac{{}^{2n}C_3}{{}^nC_3} = \dfrac{2n(2n-1)(2n-2)}{n(n-1)(n-2)} = \dfrac{4(2n-1)}{n-2}\). (i) \(=12 \Rightarrow 4(2n-1) = 12(n-2) \Rightarrow 8n-4 = 12n-24 \Rightarrow n = 5\). (ii) \(=11 \Rightarrow 8n-4 = 11n-22 \Rightarrow n = 6\).
Q3. How many chords can be drawn through 21 points on a circle?
A chord joins 2 points: \({}^{21}C_2 = 210\).
Q4. In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls?
\({}^5C_3 \cdot {}^4C_3 = 10 \cdot 4 = 40\).
Q5. Find the number of ways of selecting 9 balls from 6 red, 5 white and 5 blue balls if each selection consists of 3 balls of each colour.
\({}^6C_3 \cdot {}^5C_3 \cdot {}^5C_3 = 20 \cdot 10 \cdot 10 = 2000\).
Q6. Determine the number of 5-card combinations out of a deck of 52 cards if each selection is to contain exactly one ace.
\({}^4C_1 \cdot {}^{48}C_4 = 4 \cdot 194580 = 778320\).
Q7. In how many ways can one select a cricket team of eleven from 17 players in which only 5 players can bowl if each team must include exactly 4 bowlers?
\({}^5C_4 \cdot {}^{12}C_7 = 5 \cdot 792 = 3960\).
Q8. A bag contains 5 black and 6 red balls. Determine the number of ways in which 2 black and 3 red balls can be selected.
\({}^5C_2 \cdot {}^6C_3 = 10 \cdot 20 = 200\).
Q9. In how many ways can a student choose a programme of 5 courses if 9 courses are available and 2 specific courses are compulsory?
2 compulsory fixed; choose 3 more from 7: \({}^7C_3 = 35\).
Miscellaneous Examples
Example 17. How many words, with or without meaning, each of 2 vowels and 3 consonants, can be formed from the letters of the word DAUGHTER?
Vowels: A, U, E (3). Consonants: D, G, H, T, R (5). Select 2 vowels: \({}^3C_2 = 3\). Select 3 consonants: \({}^5C_3 = 10\). Arrange 5 letters: \(5! = 120\). Total \(= 3 \cdot 10 \cdot 120 = 3600\).
Example 18. A group consists of 4 girls and 7 boys. In how many ways can a team of 5 members be selected if the team has (i) no girl, (ii) at least one boy and one girl, (iii) at least three girls?
(i) All 5 boys: \({}^7C_5 = 21\).
(ii) Total \({}^{11}C_5 = 462\). Remove all-boy (21) and all-girl (0, since only 4 girls so no 5-girl team = 0). At least one boy and one girl = \(462 - 21 - 0 = 441\).
(iii) At least 3 girls = 3G+2B + 4G+1B = \({}^4C_3 {}^7C_2 + {}^4C_4 {}^7C_1 = 4 \cdot 21 + 1 \cdot 7 = 91\).
(ii) Total \({}^{11}C_5 = 462\). Remove all-boy (21) and all-girl (0, since only 4 girls so no 5-girl team = 0). At least one boy and one girl = \(462 - 21 - 0 = 441\).
(iii) At least 3 girls = 3G+2B + 4G+1B = \({}^4C_3 {}^7C_2 + {}^4C_4 {}^7C_1 = 4 \cdot 21 + 1 \cdot 7 = 91\).
Example 19. Find the number of arrangements of the letters of the word INDEPENDENCE. In how many of these (iv) do the words begin with I and end in P?
INDEPENDENCE: I(1), N(3), D(2), E(4), P(1), C(1); total 12. All arrangements \(= \dfrac{12!}{3!\,2!\,4!} = 1663200\). Words with I first and P last: fix those; middle 10 letters are N(3), D(2), E(4), C(1): \(\dfrac{10!}{3!\,2!\,4!} = 12600\).
Example 20. A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of (i) exactly 3 girls, (ii) at least 3 girls, (iii) at most 3 girls?
(i) 3G + 4B: \({}^4C_3 \cdot {}^9C_4 = 4 \cdot 126 = 504\).
(ii) At least 3 girls = 3G+4B + 4G+3B = \(504 + 1 \cdot 84 = 588\).
(iii) At most 3 girls = 0G+7B + 1G+6B + 2G+5B + 3G+4B = \(36 + 4 \cdot 84 + 6 \cdot 126 + 504 = 36 + 336 + 756 + 504 = 1632\).
(ii) At least 3 girls = 3G+4B + 4G+3B = \(504 + 1 \cdot 84 = 588\).
(iii) At most 3 girls = 0G+7B + 1G+6B + 2G+5B + 3G+4B = \(36 + 4 \cdot 84 + 6 \cdot 126 + 504 = 36 + 336 + 756 + 504 = 1632\).
Miscellaneous Exercise — Selected
Q1. How many words, with or without meaning, each of 3 vowels and 2 consonants, can be formed from the letters of the word INVOLUTE?
INVOLUTE has vowels I, O, U, E (4) and consonants N, V, L, T (4). \({}^4C_3 \cdot {}^4C_2 \cdot 5! = 4 \cdot 6 \cdot 120 = 2880\).
Q2. How many 4-digit numbers can be formed using the digits 1, 2, 3, 4, 5, 6, 7 if no digit is repeated and the number is divisible by 5?
Last digit must be 5 (only). Remaining 3 places from 6 remaining digits: \({}^6P_3 = 120\).
Q3. How many 6-digit numbers can be formed from the digits 0, 1, 3, 5, 7 and 9 which are divisible by 10 and no digit is repeated?
Units place must be 0. Remaining 5 places from digits {1,3,5,7,9}: \(5! = 120\).
Q4. In an examination, a question paper consists of 12 questions divided into two parts — Part I and Part II, containing 5 and 7 questions respectively. A student must attempt 8 questions in all, selecting at least 3 from each part. In how many ways can the student select the questions?
Cases (I, II): (3,5), (4,4), (5,3). \({}^5C_3 {}^7C_5 + {}^5C_4 {}^7C_4 + {}^5C_5 {}^7C_3 = 10\cdot 21 + 5 \cdot 35 + 1 \cdot 35 = 210 + 175 + 35 = 420\).
Q5. Determine the number of 5-card combinations out of a deck of 52 cards if there is exactly one ace in each combination.
\({}^4C_1 \cdot {}^{48}C_4 = 4 \cdot 194580 = 778320\).
Q6. The number lock of a suitcase has 4 wheels, each labelled with 10 digits 0–9. The lock opens with a specific sequence of 4 digits with no repetition. The first digit is 7. How many sequences must be tried to guarantee opening?
Remaining 3 wheels use 9 remaining digits without repetition: \({}^9P_3 = 9 \cdot 8 \cdot 7 = 504\) sequences.
Q7. A committee of 7 is to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of at least 3 girls?
See Ex. 20(ii): \(588\).
Q8. If the different permutations of all the letters of the word EXAMINATION are listed as in a dictionary, how many words are there in this list before the first word starting with E?
EXAMINATION letters: A(2), E(1), I(2), M(1), N(2), O(1), T(1), X(1); total 11. Words before those starting with E are those starting with A. Fix A first; arrange remaining 10 letters A(1), E, I(2), M, N(2), O, T, X: \(\dfrac{10!}{2!\,2!} = 907200\).
Competency-Based Questions — Selection in Decisions
A state education board must constitute an 8-member exam-reform committee from 6 principals, 5 teachers, and 4 student representatives. Several composition rules are under debate in the board meeting.
Q1. How many 8-member committees are possible with no restriction on composition?
L3 ApplyTotal members available \(= 6+5+4 = 15\). \({}^{15}C_8 = 6435\).
Q2. How many committees contain exactly 2 student representatives and at least 3 principals?
L4 Analyse2 students fixed (\({}^4C_2=6\)). Remaining 6 must come from 6 principals + 5 teachers with ≥3 principals. Cases (P,T): (3,3), (4,2), (5,1), (6,0). Counts: \({}^6C_3{}^5C_3 + {}^6C_4{}^5C_2 + {}^6C_5{}^5C_1 + {}^6C_6{}^5C_0 = 20\cdot 10 + 15\cdot 10 + 6\cdot 5 + 1 = 200+150+30+1 = 381\). Total \(= 6 \cdot 381 = 2286\).
Q3. A board member argues "enforcing at least 1 member from each of the three categories still leaves too few options". Evaluate: is this claim correct? Compute the count.
L5 EvaluateUse complement: "at least 1 of each" = total − (those missing principals) − (missing teachers) − (missing students) + (missing both of any two categories). Missing P: choose 8 from 9 (T+S) = \({}^9C_8 = 9\). Missing T: from 10 = \({}^{10}C_8 = 45\). Missing S: from 11 = \({}^{11}C_8 = 165\). Pairwise missing: missing P&T ⇒ all 8 from 4 students (impossible, 0); missing P&S ⇒ from 5 teachers (0); missing T&S ⇒ from 6 principals (0). Inclusion-exclusion: \(6435 - 9 - 45 - 165 + 0 = 6216\). 6216 is a very large number — the claim is not correct; there are ample options.
Q4. Design a composition rule involving all three categories that yields exactly 900 possible committees. Justify using \({}^nC_r\).
L6 CreateTry (P,T,S) = (3,3,2): \({}^6C_3 {}^5C_3 {}^4C_2 = 20 \cdot 10 \cdot 6 = 1200\) (too many). Try (4,2,2): \({}^6C_4 {}^5C_2 {}^4C_2 = 15 \cdot 10 \cdot 6 = 900\) ✓. Rule: "Exactly 4 principals, exactly 2 teachers, exactly 2 student representatives." Count = 900.
Assertion–Reason Questions
Assertion (A): \({}^nC_r = {}^nC_{n-r}\).
Reason (R): Choosing \(r\) items to include is equivalent to choosing the other \(n-r\) items to exclude.
Reason (R): Choosing \(r\) items to include is equivalent to choosing the other \(n-r\) items to exclude.
Answer: A. The bijection between selections and their complements is exactly why A holds.
Assertion (A): The number of diagonals of an \(n\)-gon (polygon with \(n\) sides) is \({}^nC_2\).
Reason (R): A diagonal joins any two vertices, and there are \({}^nC_2\) pairs of vertices.
Reason (R): A diagonal joins any two vertices, and there are \({}^nC_2\) pairs of vertices.
Answer: D. Number of diagonals is \({}^nC_2 - n\) (subtract the \(n\) sides). R is true for pair-counting but mis-applied in A.
Assertion (A): The total number of subsets of a set with \(n\) elements equals \(2^n\).
Reason (R): \(\sum_{r=0}^{n} {}^nC_r = 2^n\), since each element independently either belongs to a subset or not.
Reason (R): \(\sum_{r=0}^{n} {}^nC_r = 2^n\), since each element independently either belongs to a subset or not.
Answer: A. Both true; R is the standard double-counting justification of A.
Chapter 6 — Summary at a Glance
- Fundamental Principle of Counting: Independent events with \(m\) and \(n\) outcomes combine into \(m \times n\) outcomes (generalises to any number of stages).
- Factorial: \(n! = 1 \cdot 2 \cdots n\); \(0! = 1\); \(n! = n(n-1)!\).
- Permutations (order matters): \({}^nP_r = \dfrac{n!}{(n-r)!}\). With all \(n\) distinct used: \(n!\). With \(p_1, p_2, \ldots, p_k\) identical groups: \(\dfrac{n!}{p_1!\,p_2!\cdots p_k!}\). With repetition allowed: \(n^r\).
- Combinations (order irrelevant): \({}^nC_r = \dfrac{n!}{r!(n-r)!}\).
- Key Properties: \({}^nC_r = {}^nC_{n-r}\); Pascal: \({}^nC_r + {}^nC_{r-1} = {}^{n+1}C_r\); Sum: \(\sum_{r=0}^{n} {}^nC_r = 2^n\); Corollary: \({}^nC_a = {}^nC_b \Rightarrow a=b\) or \(a+b=n\).
- Problem-solving template: (1) Read — does order matter? Yes → permutation; No → combination. (2) Are there restrictions (together/apart, fixed position)? Use block/tie or subtract-from-total. (3) Break into cases with "at least / at most" and add. (4) Multiply independent stages; add mutually exclusive cases.
Frequently Asked Questions — Permutations and Combinations
What is Part 3 — Combinations nCr, Properties & Miscellaneous Exercises | Class 11 Maths | MyAiSchool in NCERT Class 11 Mathematics?
Part 3 — Combinations nCr, Properties & Miscellaneous Exercises | 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.
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To solve problems on Part 3 — Combinations nCr, Properties & Miscellaneous Exercises | 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 3 — Combinations nCr, Properties & Miscellaneous Exercises | Class 11 Maths | MyAiSchool important for the Class 11 board exam?
Part 3 — Combinations nCr, Properties & Miscellaneous Exercises | 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 3 — Combinations nCr, Properties & Miscellaneous Exercises | Class 11 Maths | MyAiSchool?
Common mistakes in Part 3 — Combinations nCr, Properties & Miscellaneous Exercises | 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.
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End-of-chapter NCERT exercises for Part 3 — Combinations nCr, Properties & Miscellaneous Exercises | 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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