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Class 11 Maths Notes

Permutations and Combinations Class 11 Notes

Complete, exam-ready notes on permutations and combinations: the fundamental principle of counting, factorials, arrangements (permutations) and selections (combinations), the relation between nPr and nCr, circular permutations, and arrangements of identical objects — written for CBSE boards and JEE revision.

Class11SubjectMathematicsCoversCBSE · JEE

Written byDeep Narayan· Science & Mathematics EducatorReviewed byPushpanjali

What is the core idea of this chapter?

Permutations count ordered arrangements and combinations count unordered selections — both built on the fundamental principle of counting and factorials.

Fundamental Principle of Counting

Multiplication principle

If one task can be done in m ways and a second independent task can be done in n ways, the two tasks together can be done in m × n ways. This extends to any number of sequential tasks.

Addition principle

If one task can be done in m ways and a different, mutually exclusive task can be done in n ways, exactly one of the two tasks can be done in m + n ways.

Multiplication vs addition

Use multiplication when tasks happen in sequence (AND). Use addition when tasks are alternatives (OR). Choosing between them is the first decision in every counting problem.

Factorials and Permutations

n!=n×(n−1)×(n−2)×⋯×2×1,0!=1n! = n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1, \quad 0! = 1
Factorial

Permutation

nPr=n!(n−r)!^nP_r = \frac{n!}{(n-r)!}

A permutation is an arrangement of r objects chosen from n distinct objects where order matters. The number of such arrangements is nPr=n!(n−r)!^nP_r = \frac{n!}{(n-r)!}.

  • Permutations with repetition allowed: each of the r positions has n choices, giving n^r arrangements.
  • All n objects arranged in a row: n! permutations.
  • Arranging n objects in a circle: (n − 1)! circular permutations (fixing one object eliminates rotational duplicates).

Permutations of Objects Not All Distinct

n!p! q! r! ⋯\frac{n!}{p! \, q! \, r! \, \cdots}
Permutations with identical objects

When n objects include p of one kind, q of another, r of a third, and so on, the number of distinct arrangements is n!p! q! r! ⋯\frac{n!}{p! \, q! \, r! \, \cdots}. For example, the letters of MISSISSIPPI (M=1, I=4, S=4, P=2) give 11!1! 4! 4! 2!=34650\frac{11!}{1! \, 4! \, 4! \, 2!} = 34650 distinct arrangements.

Circular Permutations

Circular permutation

In a circular arrangement, rotations of the same order are considered identical. Fixing one object's position eliminates rotational symmetry, leaving (n − 1)! arrangements of the remaining n − 1 objects.

  • If clockwise and anticlockwise order are both the same (e.g., a necklace), the count is (n − 1)!/2.
  • If direction matters (e.g., people seated around a table with a distinguished side), the count is (n − 1)!.
  • Fixing one object is the standard technique — it removes all rotational duplicates.

Combinations

Combination

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

A combination is a selection of r objects from n distinct objects where order does not matter. The number of such selections is nCr=n!r! (n−r)!^nC_r = \frac{n!}{r! \, (n-r)!}.

  • Relation between permutations and combinations:
  • nPr=r!×nCr^nP_r = r! \times ^nC_r
  • .
  • Symmetry property:
  • nCr=nCn−r^nC_r = ^nC_{n-r}
  • .
  • Pascal's identity:
  • nCr+nCr−1=n+1Cr^nC_r + ^nC_{r-1} = ^{n+1}C_r
  • .
  • Sum of all combinations:
  • ∑r=0nnCr=2n\sum_{r=0}^{n} {^nC_r} = 2^n
  • .

Key Properties and Relations

  • nPr = r! · nCr — permutations count order, combinations do not.
  • nCr = nC(n − r) — choosing r to include equals choosing n − r to exclude.
  • nCr + nC(r − 1) = (n + 1)Cr — Pascal's recurrence, the basis of Pascal's triangle.
  • For r objects taken one at a time with repetition from n objects: n^r ways.
  • When all n objects are taken (r = n): nPn = nCn = 1 arrangement/selection.

Order matters or not?

The word 'arrange' or 'order' signals a permutation; the word 'select', 'choose' or 'group' signals a combination. Misreading this is the most common mistake.

Solved Examples

Example: In how many ways can a committee of 4 men and 3 women be chosen from 7 men and 5 women?

Solution: Choose 4 men from 7: 7C4=35^7C_4 = 35. Choose 3 women from 5: 5C3=10^5C_3 = 10. By the multiplication principle, total = 35 × 10 = 350 ways.

Example: Find the number of permutations of the letters of the word ARRANGE.

Solution: ARRANGE has 7 letters with A repeated 2 times and R repeated 2 times. Distinct arrangements = 7!2! 2!=50404=1260\frac{7!}{2! \, 2!} = \frac{5040}{4} = 1260.

Revision

Key formulas at a glance

Memorise these before attempting numericals — most exam questions hinge on one of them.

Factorial

n!=n×(n−1)!n! = n \times (n-1)!

Permutations

nPr=n!(n−r)!^nP_r = \frac{n!}{(n-r)!}

Combinations

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

nPr in terms of nCr

nPr=r!×nCr^nP_r = r! \times ^nC_r

Symmetry of combinations

nCr=nCn−r^nC_r = ^nC_{n-r}

Pascal's identity

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

Identical objects

n!p! q! r! ⋯\frac{n!}{p! \, q! \, r! \, \cdots}

Exam tips

How this chapter is asked

Where this topic appears in CBSE, JEE Main and NEET papers.

  • Distinguish between 'arrange' (permutation) and 'select' (combination) — this decides whether order matters.
  • Circular permutation of n distinct objects = (n − 1)!; halve it if clockwise = anticlockwise.
  • nC0 + nC1 + ... + nCn = 2^n — the total number of subsets of an n-element set.
  • nCr = nC(n − r) lets you compute the smaller value when r > n/2.
  • With identical objects, divide n! by the factorials of the repetition counts.
  • Repetition allowed: n objects taken r at a time gives n^r arrangements.
  • Pascal's identity nCr + nC(r−1) = (n+1)Cr is useful for quick computation and proofs.

FAQ

Common questions

What is the difference between permutations and combinations?

Permutations count arrangements where order matters (e.g. ABC ≠ BAC); combinations count selections where order does not matter (e.g. {A,B,C} = {B,A,C}).

How do you calculate circular permutations?

Fix one object to remove rotational symmetry, then arrange the remaining (n − 1) objects in a line: (n − 1)! arrangements. Halve it if clockwise and anticlockwise are considered the same.

When do we use permutations with repetition?

When each of the r positions can be filled independently by any of the n objects (e.g. forming a 3-digit number where digits can repeat), the count is n^r.

What is Pascal's identity?

nCr + nC(r − 1) = (n + 1)Cr. It connects adjacent entries in Pascal's triangle and is useful for simplifying sums and proving combinatorial identities.

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