Class 11 Maths · Original EZT practice
Permutations and Combinations (Class 11 Chapter 6): summary and practice
Permutations and Combinations turns counting into formulas: the multiplication and addition principles, factorial notation, nPr when order matters, nCr when only selection matters, and the repeated-letter fixes in between. All questions are original EZT tutor practice following the chapter's own section order.
6 original worked examples, 6 fresh self-check questions and 4 mixed questions, with explained answers. These are not copied textbook exercises, an official chapter index or a full answer key. No affiliation with NCERT is claimed.
Official sources for topic and edition context, checked 21 October 2026:
1. The fundamental principle of counting
If one job can be done in m ways and another in n ways, doing both in sequence gives m x n ways; doing one OR the other gives m + n. Every formula in the chapter is these two sentences wearing algebra.
2. Factorial notation
n! is the product of the first n natural numbers, the count of ways to line up n distinct objects. The convention 0! = 1 keeps every formula working at the edges.
3. Permutations
Arranging r out of n distinct objects in order gives nPr = n! / (n - r)! ways. When some objects repeat, divide by the arrangements of the repeats: n! / (p! q! ...) - the APPLE and BANANA problems live here (6.3.4).
4. Combinations
Choosing r out of n objects when order does not matter gives nCr = n! / (r! (n - r)!) - each selection is counted once because the r! orderings are divided out. The symmetry nCr = nC(n - r) halves the work.
5. Order matters or not
The chapter's real skill is picking the right tool: arrangements, rankings and sequences are permutations; committees, teams and selections are combinations. Ask one question first - does swapping two picked items make a different outcome?
Choose, solve, check, then try a fresh question.
Before calculating, read the method prompt and name the condition that makes it useful. Write your own attempt. Open the steps only when you need them, and use the check to catch a wrong sign, unit or interpretation. Finish with the fresh question before reading its answer.
Compute a permutation
Choose the method: nPr = n! / (n - r)!.
EZT practice question
Find 5P3.
Show the worked solution
- 5P3 = 5! / 2!.
- = 5 x 4 x 3.
Answer: 60.
Check: Three ordered slots filled from five distinct items: 5 choices, then 4, then 3 - the multiplication principle directly.
Watch for this: Do not run the product all the way to 1 - nPr stops after r factors.
Now try it yourself
7P2?
Check your answer
42.
7 x 6 = 42.
Compute a combination
Choose the method: nCr = n! / (r! (n - r)!).
EZT practice question
Find 7C3.
Show the worked solution
- 7C3 = 7! / (3! 4!).
- = (7 x 6 x 5) / (3 x 2 x 1).
Answer: 35.
Check: 7P3 = 210 ordered arrangements; dividing by the 3! = 6 orderings of each trio leaves 35 selections.
Watch for this: Forgetting the r! in the denominator turns a selection count into an arrangement count.
Now try it yourself
9C2?
Check your answer
36.
(9 x 8) / 2 = 36.
Count a committee
Choose the method: Selections use nCr - a committee has no internal order.
EZT practice question
A committee of 3 is to be chosen from 12 people. How many committees are possible?
Show the worked solution
- Order inside the committee does not matter.
- 12C3 = (12 x 11 x 10) / (3 x 2 x 1).
Answer: 220 committees.
Check: Symmetry check: 12C3 = 12C9, and choosing the 3 who serve equals choosing the 9 who stay.
Watch for this: If the posts were distinct (president, secretary, treasurer), it would be a permutation - same people, different answer.
Now try it yourself
8C3?
Check your answer
56.
(8 x 7 x 6) / 6 = 56.
Work with factorials
Choose the method: n! = n x (n - 1) x ... x 1, with 0! = 1.
EZT practice question
Find 8!.
Show the worked solution
- Build up: 5! = 120, 6! = 720, 7! = 5040.
- 8! = 8 x 5040.
Answer: 8! = 40320.
Check: Factorials grow brutally fast - 10! is already over 3.6 million, which is why counting needs formulas rather than lists.
Watch for this: 0! = 1, not 0 - the convention keeps n! / (n - r)! valid when r = n.
Now try it yourself
6!?
Check your answer
720.
6 x 5 x 4 x 3 x 2 x 1 = 720.
Arrange words with repeats
Choose the method: Divide n! by the factorial of each repeat count.
EZT practice question
How many distinct arrangements does the word APPLE have?
Show the worked solution
- 5 letters, P repeats twice.
- Arrangements = 5! / 2!.
- = 120 / 2.
Answer: 60.
Check: The two P's can swap without making a new word, so every raw arrangement was counted twice.
Watch for this: Only repeated letters get divided out - A, L and E appear once and cost nothing.
Now try it yourself
Arrangements of BANANA?
Check your answer
60.
6! / (3! 2!) = 720 / 12 = 60, dividing for three A's and two N's.
Apply the multiplication principle
Choose the method: Independent choices multiply.
EZT practice question
A student picks one of 4 shirts, one of 3 pairs of trousers and one of 2 pairs of shoes. How many outfits are possible?
Show the worked solution
- Three independent choices.
- 4 x 3 x 2.
Answer: 24 outfits.
Check: Had the question asked for a shirt OR shoes, the addition principle would give 4 + 2 = 6.
Watch for this: AND multiplies, OR adds - mixing the two is the chapter's most common slip.
Now try it yourself
A code uses one of 3 letters followed by one of 4 digits. How many codes?
Check your answer
12.
3 x 4 = 12.
Try a mixed set
Choose a method or explanation before opening each answer.
1. nPr?
Show answer and reason
n! / (n - r)!.
Order matters - arrangements.
2. nCr?
Show answer and reason
n! / (r! (n - r)!).
Order does not matter - selections.
3. Repeated letters?
Show answer and reason
Divide n! by each repeat's factorial.
Swapping identical items changes nothing.
4. 0!?
Show answer and reason
1.
Keeps every formula consistent at r = n.
Try it yourself
Work these on paper before checking any notes. Answers are deliberately not shown here.
- Find 7P2.
- Find 9C2.
- How many distinct arrangements does BANANA have?
Know the rule but lose track of the steps?
Bring your book edition, your school topic list and a question you attempted. Ask a tutor to explain the choice of method and test it with a different problem. Agree the lesson topics, schedule and fee before starting.
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Questions about this practice page
- When do I use nPr and when nCr?
- Ask whether swapping two chosen items gives a different outcome. Arrangements, ranks and sequences: yes, use nPr. Committees, teams and selections: no, use nCr. The same objects can need either formula depending on the question.
- Why is 0! equal to 1?
- There is exactly one way to arrange zero objects - do nothing. Defining 0! = 1 keeps nPr = n! / (n - r)! and nCr working when r = n, so the formulas never need exceptions.
- What does the fundamental counting principle say?
- Two sentences: independent choices done together MULTIPLY (m ways then n ways gives m x n), and mutually exclusive choices ADD (m ways or n ways gives m + n). Every permutation and combination formula is built from these.
- How do I handle repeated letters in a word?
- Count the arrangements as if all letters were distinct (n!), then divide by the factorial of each letter's repeat count - those swaps never made a new word. APPLE gives 5! / 2! = 60; BANANA gives 6! / (3! 2!) = 60.
- Are these the NCERT exercise questions for Permutations and Combinations (Class 11)?
- No. EZT pages are original practice following the chapter's section order - use them to prepare for the textbook exercises, not as a substitute for your book.