Permutations & Combinations

Concept Core

Arrangement vs selection — and how to count both correctly.

The Fundamental Principle of Counting

If one event can occur in m ways and a second, independent event in n ways, they can occur together in m × n ways (Multiplication Principle). If they are mutually exclusive, they can occur in m + n ways (Addition Principle). All counting formulas stem from these two principles.

Permutations — Order Matters

Number of ways to arrange r objects out of n distinct objects where order matters:

ⁿPᵣ = n! / (n−r)! (0 ≤ r ≤ n)

All n objects: ⁿPₙ = n!

When repetition allowed: nʳ arrangements (each of r positions has n choices).

Combinations — Order Doesn't Matter

Number of ways to select (not arrange) r objects from n distinct objects:

ⁿCᵣ = n! / (r!(n−r)!) = ⁿPᵣ / r!

Key identity: ⁿCᵣ = ⁿCₙ₋ᵣ (selecting r is same as rejecting n−r)

ⁿC₀ = ⁿCₙ = 1 always.

Circular Permutations

Arranging n objects in a circle: (n−1)! ways (one position is fixed to avoid counting rotations).

Necklace/bracelet: (n−1)!/2 (flips are also identical).

Eg: 5 people at a round table = 4! = 24 ways.

Permutations with Repetition

Arranging n objects where p are alike of one kind, q alike of another, etc.:

n! / (p! × q! × r! × ...)

Eg: MISSISSIPPI has 11 letters (4 S, 4 I, 2 P, 1 M): 11!/(4!4!2!1!) arrangements.

Pascal's Triangle Identities

Sum identity: ⁿCᵣ + ⁿCᵣ₋₁ = ⁿ⁺¹Cᵣ (Pascal's rule)

Row sum: ΣⁿCᵣ = 2ⁿ (total subsets of n elements)

Alternating sum: Σ(−1)ʳ ⁿCᵣ = 0 (for n ≥ 1)

Symmetric: ⁿCᵣ = ⁿCₙ₋ᵣ

Strategic Framework — Which Formula to Use?

→ Order matters, no repetition: ⁿPᵣ = n!/(n−r)!

→ Order matters, with repetition: nʳ

→ Order doesn't matter: ⁿCᵣ

→ In a circle: (n−1)!

→ With alike objects: n!/(p!q!...)

The key question: does changing the order give a DIFFERENT outcome? If YES → permutation. If NO → combination.

Formula Vault

All counting formulas in one place.

Permutation

ⁿPᵣ = n!/(n−r)!

r items from n, order matters

Combination

ⁿCᵣ = n!/(r!(n−r)!)

r items from n, order ignored

All Permutations

ⁿPₙ = n!

All n objects arranged

With Repetition

nʳ arrangements

r positions, each n choices

Alike Objects

n!/(p!q!r!...)

p alike type1, q alike type2...

Circular Perm.

(n−1)!

n objects in circle, 1 fixed

Necklace

(n−1)!/2

Flipping is same arrangement

Symmetric Property

ⁿCᵣ = ⁿCₙ₋ᵣ

Select r = reject n−r

Pascal's Rule

ⁿCᵣ + ⁿCᵣ₋₁ = ⁿ⁺¹Cᵣ

Foundation of Pascal's triangle

Total Subsets

ΣⁿCᵣ = 2ⁿ

Including empty set

Worked Examples

5 problems — selection, arrangement, circular, and EAPCET traps.

EasyIn how many ways can 4 students be chosen from 10 for a team?▾

Choose 4 students from 10 for a team. Order doesn't matter.

Order doesn't matter → use Combination.

¹⁰C₄ = 10!/(4!×6!) = (10×9×8×7)/(4×3×2×1) = 5040/24 = 210

✓ 210 ways

EasyHow many 3-letter words from EAPCET (all distinct letters chosen)?▾

From the 6 distinct letters of EAPCET (E,A,P,C,T — treating both E's as one for now, 6 distinct available), form 3-letter words where order matters.

Order matters (different arrangements = different words) → ⁿPᵣ

From 6 distinct letters, 3 at a time: ⁶P₃ = 6!/(6−3)! = 6!/3! = 720/6 = 120

✓ 120 3-letter arrangements

MediumIn how many ways can 5 people sit at a round table?▾

Arrange 5 people at a circular table.

For circular arrangements, fix one person to remove rotational duplicates.

Remaining 4 people can be arranged in 4! = 24 ways

Formula confirms: (n−1)! = (5−1)! = 4! = 24

✓ 24 circular arrangements

EAPCET LevelA committee of 3 men and 2 women from 6 men and 4 women — how many ways?▾

Select a committee of 3 men and 2 women from 6 men and 4 women.

Men selection (order doesn't matter): ⁶C₃ = 20

Women selection: ⁴C₂ = 6

Both selections are independent → multiply: 20 × 6 = 120

✓ 120 committees

Trap QuestionHow many ways to arrange letters of MISSISSIPPI?▾

Count distinct arrangements of the letters in MISSISSIPPI. ⚠️ Students forget repeated letters.

Count letters: M=1, I=4, S=4, P=2. Total = 11 letters.

The trap: Using 11! without dividing for repeats counts the same arrangement multiple times.

Correct formula: n!/(p!q!r!) = 11!/(1!×4!×4!×2!)

= 39916800/(1×24×24×2) = 39916800/1152 = 34650

✓ 34,650 distinct arrangements

Mistake DNA

4 errors from EAPCET distractor analysis in P&C.

🔀

Using Permutation When Combination is Needed

'Select a team of 4' uses ⁿCᵣ not ⁿPᵣ. Order within a team doesn't create a new team.

❌ Wrong

Team of 4 from 10: ¹⁰P₄ = 5040 ✗ (treats orderings as different teams)

✓ Correct

¹⁰C₄ = 210 ✓ Same team in any order = 1 selection

Ask: does changing the order of selected items give a different result? Committee/team/group → Combination. Password/rank/queue → Permutation.

🌀

Circular Permutation Using n! Instead of (n−1)!

Not fixing one element means all rotations of the same arrangement are counted separately.

❌ Wrong

5 people round table: 5! = 120 ✗ (overcounts by factor 5)

✓ Correct

(5−1)! = 4! = 24 ✓ Fix one person; arrange remaining 4

In a circle, rotating everyone by one seat gives the SAME arrangement. Fixing one person eliminates this overcounting.

🔢

Not Dividing by Repeated Elements in Word Arrangements

When letters repeat, many arrangements are identical. Forgetting to divide massively overcounts.

❌ Wrong

MISSISSIPPI arrangements: 11! = 39916800 ✗ (ignores 4I,4S,2P)

✓ Correct

11!/(4!4!2!) = 34650 ✓ Divide by factorial of each repeated group

Each group of p identical letters can be rearranged p! ways without creating new words. Divide by p! to eliminate these duplicates.

➕

Adding Instead of Multiplying for Sequential Independent Events

When TWO independent choices are made simultaneously, multiply, not add.

❌ Wrong

Choose 1 from 5 boys AND 1 from 4 girls: 5 + 4 = 9 ✗ (these are both required)

✓ Correct

5 × 4 = 20 ✓ Multiply when BOTH events must occur

Multiplication Principle: if event A has m outcomes AND event B has n outcomes (both must happen), total = m×n. Use addition only when events are mutually exclusive (either A OR B).

Chapter Intelligence

P&C is a direct prerequisite for Probability — master both together.

EAPCET Weightage (2019–2024)

Combinations (selection)

Permutations (arrangement)

Circular permutation

Word arrangements (alike)

Total subsets / Pascal's

High-Yield PYQ Patterns

Select committee with conditionsCircular seating of n peopleArrange letters of a wordPasswords with restrictionsWays to distribute identical objectsⁿCᵣ = ⁿCₛ → r+s = n

Exam Strategy

Always determine: Does order matter? Yes → Permutation. No → Combination. This one check prevents the most common error.
For circular problems: answer = (n−1)! for people; (n−1)!/2 for necklaces/chains (two sides are same).
Conditional selection (at least one woman, exactly 2 men, etc.): enumerate cases and add, using Multiplication Principle within each case.
P&C directly enables Probability. If you know P&C cold, Probability becomes straightforward — favourable/total outcomes.
ⁿCᵣ = ⁿCₛ implies r = s or r+s = n. This identity appears frequently in EAPCET as "find n given ⁿCr = ⁿCₛ" questions.

Concept Core

Formula Vault

Worked Examples

Mistake DNA

Chapter Intelligence

💡 Suggestions & Feedback

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