Permutations & Combinations

Counting problems test careful case analysis rather than formula recall. Advanced favours arrangements with restrictions, distributions (identical/distinct objects and boxes), the inclusion–exclusion principle, and derangements. Framing the problem correctly is 90% of the work.

Key results

• Basics: nPr = n!/(n−r)!, nCr = n!/(r!(n−r)!); circular arrangements (n−1)!.
• Identical objects: arrangements of n with repeats = n!/(p!q!...).
• Distribution: distinct objects into distinct boxes rⁿ; identical objects into distinct boxes = C(n+r−1, r−1) (stars and bars).
• Inclusion–exclusion: |A∪B∪C| = Σ|A| − Σ|A∩B| + |A∩B∩C|.
• Derangements: Dₙ = n!·Σ(−1)^k/k! (no element in its original place).

Task	Count
Arrange n distinct in a row	n!
Circular arrangement	(n−1)!
Choose r from n	nCr
Identical items in boxes	C(n+r−1, r−1)

Exam Tricks & Tips

• 🎯 Use stars and bars for identical objects into distinct boxes: C(n+r−1, r−1) — the standard non-negative-integer-solution count.
• 🎯 Inclusion–exclusion handles "at least one" restrictions — count the complement (none) and subtract, or alternate over intersections.
• 🎯 Fix one person in circular arrangements to remove rotational symmetry — giving (n−1)!; divide by 2 further if reflections coincide (necklaces).
• 🎯 The number of ways to distribute so nobody keeps their own item is a derangement Dₙ ≈ n!/e — memorise D3=2, D4=9, D5=44.
• 🎯 Separate selection from arrangement: choose first (nCr), then order (×r!) only if order matters — mixing these is the top error.
• ❌ Common mistake: double-counting in "at least" problems by adding overlapping cases instead of using inclusion–exclusion.

Expected exam pattern

1–2 questions: a restricted-arrangement or distribution problem and an inclusion–exclusion/derangement conceptual (often integer). Case-heavy and error-prone.

Quick recap

nPr order matters, nCr does not; circular (n−1)!. Identical-object distribution via stars and bars C(n+r−1, r−1). Inclusion–exclusion for "at least one"; derangements Dₙ ≈ n!/e. Always separate choosing from ordering.

Cover the answer, recall, then check. 11 cards on permutations and combinations for JEE Advanced.

Q1. Formula for nPr and nCr.
A1. nPr = n!/(n−r)!; nCr = n!/(r!(n−r)!).

Q2. Number of circular arrangements of n distinct objects.
A2. (n − 1)! (fix one to remove rotations).

Q3. Arrangements of n objects with repeats p, q, ...
A3. n!/(p! q! ...).

Q4. Number of non-negative integer solutions of x1+...+x_r = n.
A4. C(n + r − 1, r − 1) (stars and bars).

Q5. Inclusion–exclusion for three sets.
A5. |A∪B∪C| = Σ|A| − Σ|A∩B| + |A∩B∩C|.

Q6. What is a derangement?
A6. A permutation in which no element is in its original position; Dₙ = n!Σ(−1)^k/k!.

Q7. Values of D3, D4, D5.
A7. 2, 9, and 44.

Q8. Ways to distribute n distinct objects into r distinct boxes.
A8. rⁿ.

Q9. When do you multiply a combination by r!?
A9. When, after selecting r objects, their order also matters (turning selection into arrangement).

Q10. Number of ways to seat n people around a round table if two arrangements that are reflections are identical.
A10. (n − 1)!/2.

Q11. How to count "at least one" satisfying a condition?
A11. Total minus the count with "none" (complementary counting) or inclusion–exclusion.

Counting is deceptively hard: JEE Advanced problems hide subtle over- or under-counting, restrictions, and identical objects. The winning approach is a clear model (arrangement vs selection, with/without repetition) plus systematic case analysis.

Core concept: permutations count ordered arrangements, combinations count unordered selections; every problem is a careful application of the addition (OR) and multiplication (AND) principles, corrected for symmetry and identity.

Core ideas

Beginner — fundamental principles and formulas

Multiplication principle: independent choices multiply; addition principle: mutually exclusive cases add. Permutations of n distinct objects taken r: ⁿPᵣ = n!/(n−r)!. Combinations: ⁿCᵣ = n!/(r!(n−r)!). Order matters ⇒ permutation; order irrelevant ⇒ combination. Arrangements of n objects with repeats (p alike, q alike): n!/(p!q!…).

Intermediate — restrictions and standard models

• Circular permutations: (n−1)! arrangements (fix one to kill rotational symmetry); ÷2 more if reflections coincide (necklace/garland).
• Gap method (no two together): arrange the others, then place the restricted items in the gaps.
• Together (grouping): treat a block as one unit, then permute inside.
• Distribution: distributing r identical objects into n distinct boxes = C(r+n−1, n−1) (stars and bars); at least one each = C(r−1, n−1).

Advanced — inclusion–exclusion and derangements

Inclusion–exclusion counts "at least one" or "none of" conditions: |A∪B∪C| = ΣA − ΣAB + ABC. Derangements (no object in its own place): D_n = n!(1 − 1/1! + 1/2! − … + (−1)ⁿ/n!). Number of onto functions, integer solutions with bounds, and "divisible by / at least k conditions" problems all reduce to inclusion–exclusion. Selecting with at least one condition is usually easier as (total − complement). Multinomial and the "grouping into unequal/equal groups" (÷ symmetry) cases are classic traps.

Worked example

In how many ways can 5 boys and 3 girls be seated in a row so that no two girls are adjacent? Use the gap method: seat the 5 boys first in 5! = 120 ways. This creates 6 gaps (including the ends): B_B_B_B_B. Place the 3 girls in 3 of these 6 gaps, order mattering: ⁶P₃ = 6·5·4 = 120 ways. Total = 120 × 120 = 14400. The gap method automatically enforces "no two girls together" — trying to subtract the bad cases directly would be far messier.

How JEE Advanced tests this

Arrangement/selection with restrictions (together, not-together, alternate); circular and necklace permutations; distribution of identical/distinct objects (stars and bars); inclusion–exclusion for "at least/exactly/none" conditions; derangements; forming numbers with digit constraints; and geometric counting (points, lines, triangles, diagonals).

Exam tricks & shortcuts

• Order matters ⇒ P; doesn't ⇒ C. "At least one" ⇒ total − none.
• No-two-together ⇒ gap method; together ⇒ treat as a block.
• Circular: (n−1)!; identical objects: divide by the factorials of repeats.
• Identical objects into distinct boxes: C(r+n−1, n−1).

Key formulas

• ⁿPᵣ = n!/(n−r)!; ⁿCᵣ = n!/[r!(n−r)!].
• Identical items: n!/(p! q! r!); repetition allowed: nʳ.
• Circular: (n−1)!.
• ⁿCᵣ = ⁿC₍ₙ₋ᵣ₎; ΣⁿCᵣ = 2ⁿ; Pascal ⁿCᵣ + ⁿC₍ᵣ₋₁₎ = ⁿ⁺¹Cᵣ.
• Distribution/partition: dearrangements Dₙ = n!Σ(−1)ᵏ/k!.

Permutations count arrangements, combinations count selections; multiply for AND, add for OR (exclusive cases).

Worked Example

Problem: Using all the letters of the word "ARRANGE", find (a) the total number of distinct arrangements and (b) the number of arrangements in which the two R's are never together.

Solution:
"ARRANGE" has 7 letters: A appears twice, R appears twice, and N, G, E once each.

(a) Total distinct arrangements (dividing by the repeats):
Total = 7! / (2! · 2!) = 5040 / 4 = 1260.

(b) First count arrangements with the two R's together: tie the two R's into a single block "RR". Now arrange 6 units {RR, A, A, N, G, E}, where A repeats twice:
Together = 6! / 2! = 720 / 2 = 360.

Arrangements with the R's never together = total − together:
1260 − 360 = 900.

Answer: (a) 1260 distinct arrangements; (b) 900 with the R's never together.