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.
1
Order doesn't matter → use Combination.
2
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.
1
Order matters (different arrangements = different words) → ⁿPᵣ
2
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.
1
For circular arrangements, fix one person to remove rotational duplicates.
2
Remaining 4 people can be arranged in 4! = 24 ways
3
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.
1
Men selection (order doesn't matter): ⁶C₃ = 20
2
Women selection: ⁴C₂ = 6
3
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.
1
Count letters: M=1, I=4, S=4, P=2. Total = 11 letters.
2
The trap: Using 11! without dividing for repeats counts the same arrangement multiple times.
3
Correct formula: n!/(p!q!r!) = 11!/(1!×4!×4!×2!)
4
= 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)~7 Permutations (arrangement)~6 Word arrangements (alike)~3 Total subsets / Pascal's~2 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.