NMAT – Data Interpretation & Logical Reasoning
Topic: Arrangements
Short Title: Linear & Circular Arrangements with Constraints
Target Audience: NMAT aspirants preparing for the Logical Reasoning section.
1. Simple Explanation
In “arrangements” problems you are asked to find how many different ways a set of items (people, objects, seats, etc.) can be positioned, under given rules or constraints.
- Linear arrangements – items are placed in a straight line (e.g., a row of seats).
- Circular arrangements – items are placed around a circle (e.g., a round table).
- Constraints – conditions such as “X must sit next to Y”, “Z cannot sit on the leftmost seat”, “no two boys together”, etc.
The tools you’ll use:
| Tool | What it gives | Formula |
|---|---|---|
| Permutation | Number of ways to arrange n distinct items in a line | ( n! ) |
| Combination | Number of ways to choose k items from n (order doesn’t matter) | ( \binom{n}{k} = \frac{n!}{k!(n-k)!} ) |
| Circular permutation | Arrangements around a circle (rotations considered same) | ( (n-1)! ) |
| Inclusion–Exclusion | Count arrangements that satisfy multiple overlapping constraints | ( |
2. Key Points to Remember
| # | Point | Why It Matters |
|---|---|---|
| 1 | Differentiate between permutations and combinations | Most NMAT problems need order, so permutations are usually the start point. |
| 2 | Linear vs Circular | Circular arrangements divide by (n) (or fix one position) because rotations are identical. |
| 3 | Fix a position to break symmetry | In circular problems, fix one person/object at a seat to avoid over‑counting. |
| 4 | Use complement counting | When a rule forbids a condition (e.g., “A and B cannot sit together”), count total arrangements and subtract those that violate the rule. |
| 5 | Chain restrictions | Sometimes you must satisfy several conditions simultaneously; use Inclusion–Exclusion or step‑by‑step “block” methods. |
| 6 | Block method | Treat a group that must stay together as a single item, then permute the blocks and later permute inside the block. |
| 7 | Keep track of “adjacent” vs “next to” | “Next to” means directly adjacent; “adjacent” can also mean any two consecutive positions. |
| 8 | Check for identical items | If some objects are identical (e.g., 2 identical books), divide by factorial of identical count. |
| 9 | Use “stars and bars” for distributing identical items | When items are identical and you need to distribute them in slots. |
| 10 | Always double‑check constraints after each calculation | Small mis‑interpretation of a rule can give a completely wrong answer. |
3. Worked Example
Problem (NMAT style)
*Five people – A, B, C, D, and E – are to sit in a row of five seats.
Conditions:
- A must sit to the left of B.
- C cannot sit next to D.
- E must not sit in the middle seat.*
How many valid seatings are possible?
Step‑by‑Step Solution
| Step | Explanation | Calculation |
|---|---|---|
| 1. Count total linear arrangements | 5 distinct people | (5! = 120) |
| 2. Apply condition 1 (A left of B) | For any pair A & B, exactly half the arrangements have A left of B. | ( \frac{120}{2} = 60 ) |
| 3. Apply condition 3 (E not in middle) | In the 60 arrangements, 1/5 have E in middle, because each seat is equally likely for E. | (60 - \frac{60}{5} = 60 - 12 = 48) |
| 4. Apply condition 2 (C not next to D) | Count arrangements where C & D are adjacent and subtract. • Treat C & D as a block (2 orders inside). • Block + remaining 3 persons → 4 items. • Arrangements of 4 items: (4! = 24). • Inside block: 2! = 2. • So adjacent arrangements = (24 \times 2 = 48). • But we need only those that also satisfy conditions 1 & 3. • Count adjacent arrangements that already satisfy A left of B and E not in middle. |
|
| 5. Count adjacent arrangements violating C‑D rule but still meeting 1 & 3 | • Fix the block CD or DC. • Arrange the block with other 3 people (A, B, E). • Among the 4! = 24 permutations of [block, A, B, E], half have A left of B (since A & B are distinct). → 12. • Now enforce E not in middle: 12 permutations → subtract those with E in middle. • Number of permutations with E in middle: place E in middle, then arrange block, A, B in remaining 3 slots (3! = 6). • Among these 6, half have A left of B → 3. • So adjacent arrangements meeting |