Seating Arrangement – Advanced Study Notes
(Reasoning Ability – Competitive Exams – IIT‑JEE, JEE‑Main, GATE, etc.)
Goal: Master the combinatorial techniques for all kinds of seating‑arrangement problems.
Scope: Linear and circular seating, adjacency, separation, gender alternation, identical persons, and mixed constraints.
1. Foundations
| Symbol | Meaning | Example |
|---|---|---|
| (n!) | Number of permutations of (n) distinct people in a row | 4 people → (4! = 24) |
| ((n-1)!) | Number of distinct arrangements around a round table (rotational symmetry removed) | 5 people → (4! = 24) |
| (\binom{n}{k}) | Number of ways to choose (k) people from (n) | 8 choose 3 → (\binom{8}{3}=56) |
| (2^k) | Internal arrangements of (k) adjacent blocks (each block can be flipped) | 2 adjacent people → 2 ways |
Key Insight: Treat constraints as blocks or forbidden positions, then count the remaining free places.
2. Linear Seating (Row)
2.1 Unconstrained
[ \text{Ways} = n! ]
2.2 Adjacent Constraints
(a) One specified pair must sit together
Treat the pair as a single block.
- Number of objects to permute: (n-1).
- Internal permutations of the block: 2.
[ \boxed{\text{Ways} = 2,(n-1)!} ]
(b) Two specified disjoint pairs must sit together
Treat each pair as a block → two blocks + (n-4) singletons → (n-2) objects.
[ \boxed{\text{Ways} = 2^2,(n-2)!} ]
(c) (k) specified disjoint pairs sit together
[ \boxed{\text{Ways} = 2^k,(n-k)!} ]
(d) At least one of the (m) specified pairs sits together
Use Inclusion–Exclusion:
[ \begin{aligned} N_{\ge 1} &= \sum_{i=1}^{m} (-1)^{i+1} \binom{m}{i}, 2^i,(n-i)! . \end{aligned} ]
Derivation:
For each subset of (i) pairs, treat them as (i) blocks. The number of permutations of the resulting (n-i) objects is ((n-i)!). Multiply by (2^i) for internal arrangements.
Alternate signs arise because of Over‑counting when multiple pairs are simultaneously adjacent.
2.3 Separation Constraints
(a) At least (k) people between two specified persons
-
Place the two specified persons first.
They occupy 2 of the (n) seats.
The remaining (n-2) seats must be split into (2) groups of size (\ge k). -
Stars‑and‑Bars:
Let (x) be the number of seats before the first specified person, (y) the number of seats between the two, (z) the number after the second.
Constraints: (y \ge k,; x,z \ge 0,; x+y+z = n-2).Set (y' = y-k) → (x + y' + z = n-2-k).
Number of non‑negative solutions: (\binom{n-2-k+2}{2} = \binom{n-k}{2}). -
Arrange the other (n-2) people: ((n-2)!).
[ \boxed{\text{Ways} = \binom{n-k}{2},(n-2)!