NCERT Exemplar Solutions Class 11 Maths Chapter 6 Permutations and Combinations

Five of the 64 questions in the Class 11 Maths Chapter 6 Exemplar are Match the Following, and one of them cannot be matched at all. The NCERT Exemplar Solutions for Class 11 Maths Chapter 6 Permutations and Combinations on this page solve all 64 questions of the chapter, step by step, according to the 2026-27 NCERT Exemplar.

Every solution in this Collegedunia set is prepared by subject experts, based on the 2026-27 NCERT Exemplar, and checked line by line against the official answer key.

The Exemplar prints this chapter as Chapter 7, so its single long exercise is numbered Exercise 7.3 in the book's own numbering. The solutions PDF runs to 220 pages and covers every question in it.

• 64 questions solved: Short Answer, Long Answer, Objective, Fill in the Blanks, True/False and Match the Following.
• Four printed errors flagged: a wrong hint on Q14, duplicate options on Q40, and a wrong key on Q41 and Q61.
• Exam focus: useful for CBSE Boards, JEE Main, JEE Advanced and CUET.

Topics Covered in the Class 11 Maths Permutations and Combinations Exemplar

The Exemplar keeps the textbook's topic list but hides the counting rule inside a story. Most questions are one line long and still need a case split. These are the topics the 64 questions are built on.

• Fundamental principle of counting: deciding whether the stages multiply or the cases add.
• Permutations: ⁿP_r when order matters, including numbered seats and digit problems.
• Combinations: ⁿC_r when order does not matter, plus the symmetry and Pascal results.
• Alike objects: dividing by the repeats, as in n! / (p! q! r!) for words with repeated letters.
• Circular and two-sided arrangements: (n − 1)! for a round table, and why a long table is different.
• Grouping and distribution: splitting people into named groups against unnamed groups.
• Selections and subsets: 2ⁿ total selections and 2ⁿ − 1 non-empty ones.
• Counting in geometry: lines and triangles from a set of points, with a correction for collinear points.

Important: the single most useful habit in this chapter is to ask "does order matter?" before writing anything. Get that wrong and every later step is wrong, however clean the arithmetic looks.

Exercise 7.3 Question-Type Breakdown for Permutations and Combinations

All 64 questions sit inside Exercise 7.3, grouped by format. The table below shows how the exercise is split, so students can plan practice by question type rather than solving straight through.

The Objective block (Q26 to Q40) is the largest group and the one that rewards speed, so it is worth a full practice session on its own. The Match the Following block (Q60 to Q64) is unusual: each question is really four questions in one, which makes those five items worth about twenty.

Four Printed Errors in the Permutations and Combinations Exemplar Every Student Must Know

The printed Chapter 7 material has four questions that carry an error, and they are not all the same kind of error. One hint is misprinted, one option list is defective, and two keys give the wrong answer. Each is checked below against the book's own results, so read this section before self-checking.

Q14 is a misprinted hint, not a wrong answer. The box has 2 white, 3 black and 4 red balls, and 3 balls are drawn with at least one black. The book's hint reads ³C₁ × ⁶C₂ + ³C₂ × ⁶C₂ + ³C₃, which totals 91. The middle term is the misprint. The one-line check: after 2 black balls only 3 − 2 = 1 ball is left to draw, so the term must be ³C₂ × ⁶C₁ = 18, not 45. Written the book's way it would draw 4 balls, not 3. The correct total is 45 + 18 + 1 = 64, which the official key to Q50, an identical question, confirms.

Q40 has a defective option list, but the key is right. The question asks for selections of at least one green dye, at least one blue dye and any number of red dyes. The book prints option (A) as 3600 and option (D) as 3600, the same number twice, so the list is defective: two options cannot share a value in a well-posed MCQ, because neither could then be uniquely correct. The one-line check: (2⁵ − 1)(2⁴ − 1) × 2³ = 31 × 15 × 8 = 3720, which is option (B), and the key does print (B). The maths is unaffected here, only the printed choices are.

Q41: the key answers a different question. The blank reads "r = ______", but the key prints "n = 7". The one-line check: r! = ⁿP_r / ⁿC_r = 840 / 35 = 24 = 4!, so r = 4. The value n = 7 is itself correct, so the slip is in which unknown was reported, not in the working. Students should write r = 4 to answer the question as printed.

Why Q61 of the Permutations and Combinations Exemplar Cannot Be Matched

Q61 deserves its own section because it is the only question in the chapter that has no valid one-to-one matching at all. Five boys and five girls form a line, and four seating conditions must be matched to four values. The key prints (c) ↔ (iv), and that is wrong on its own terms, but the deeper problem is that the question cannot be matched however it is answered.

• The key's answer fails a direct check. "All the girls sit together" treats the five girls as one block: 6 units in 6! orders, girls inside in 5! orders, so 6! × 5! = 86,400. That is option (i). Option (iv) is 2! 5! 5! = 28,800, which is a different number.
• Two conditions share one value. Part (b), "no two girls sit together", and part (c), "all the girls sit together", both work out to 5! × 6!, so both need option (i).
• Two options share one value. Option (iii) is (5!)² + (5!)² and option (iv) is 2! 5! 5!. Since (5!)² + (5!)² = 2(5!)² = 2! 5! 5!, they are two spellings of the same number, 28,800.
• The book contradicts its own key. Option (ii) is 10! − 5! 6!, which can only mean "total minus all girls together". So the book itself is saying that "all the girls sit together" equals 5! 6!, which is option (i), not (iv).

What option (iv) actually answers. The value 2! 5! 5! = 28,800 is the correct count for a question that was never asked: "all the girls sit together and all the boys sit together". That gives two blocks in 2! orders, each block arranged internally in 5! ways. The corrected matching is (a) ↔ (iii) = 28,800, (b) ↔ (i) = 86,400, (c) ↔ (i) = 86,400 and (d) ↔ (ii) = 3,542,400. Students should solve the maths, state (c) ↔ (i), and not reproduce the key.

Key Formulas and Results Used in the Permutations and Combinations Exemplar

This chapter runs on a short list of formulas used in long combinations. Students should know these by heart before starting Exercise 7.3.

Tip: the formula students most often misapply is the last one. The collinear points do not stop making a line, they only stop making ^mC₂ different lines. So the shape is subtract the pairs, add back one line, and the "+ 1" is the part that gets dropped.

Common Mistakes Students Make in the Permutations and Combinations Exemplar

The Exemplar sets the same traps every year, and most of them are traps of reading, not of arithmetic. These are the ones that cost the most marks in this chapter.

• Selecting but not arranging. In Q30 the question says "words formed", so the 5 chosen letters must still be arranged. Stopping at ⁴C₂ × ⁵C₃ = 60 misses the × 5! and lands on the trap option.
• Dividing by the group count when the groups are named. In Q21 the groups are "two experimental and one control", so they are distinguishable and there is no division by 3!.
• Treating a long table like a round one. In Q57 the two sides are distinguishable, so never divide by 2!, and once side A is chosen side B is fixed.
• Forgetting one repeat-division. In Q44 the vowels and the consonants each need their own division by the repeats. Doing one and not the other is the standard slip.
• Missing the longer numbers. "Greater than 7000" in Q16 includes every 5-digit number: 120 of the 192 integers have five digits.
• Mishandling the leading zero. Telephone numbers may start with 0, integers may not. The correction applies only to the case that actually contains a zero.

How the Permutations and Combinations Exemplar Steps Up from the NCERT Textbook

The Exemplar adds no new topics. It asks the same ideas in a harder way, usually by burying the case split inside the wording, as the table shows.

How to Use the Permutations and Combinations Exemplar for Boards and JEE Preparation

Treat Exercise 7.3 as four sittings, not one. Finish the NCERT textbook exercises first, because the Exemplar assumes students can already choose between ⁿP_r and ⁿC_r without thinking.

That is about 7 hours for the chapter. For JEE Main and JEE Advanced, the Objective block and the grouping questions matter most, because both reward a fast, correct case split. For CBSE Boards and CUET, the Long Answer block and the Fill in the Blanks work on ⁿP_r and ⁿC_r are the better use of time.

Practice Questions for Class 11 Maths Permutations and Combinations

After reading the solutions, students should test themselves on the same question types. The card below opens a set of solved practice questions with step-by-step answers for Permutations and Combinations.

Other Resources for Class 11 Maths Permutations and Combinations

Pair the Exemplar Solutions with the textbook solutions and the notes for full chapter revision. All the Permutations and Combinations resources are linked below.

NCERT Exemplar Solutions for Other Class 11 Maths Chapters

The full Class 11 Maths Exemplar set covers 14 chapters, 735 questions and 2,625 pages of solutions. Use the table to jump to any other chapter.