Arrangements: Data Interpretation & Logical Reasoning Notes for SNAP

Arrangements – Data Interpretation & Logical Reasoning

(SNAP – Standardised National Aptitude Program)

1. Simple Explanation

An arrangement is a way of putting objects (people, books, letters, etc.) in a particular order or pattern.

Order matters – “A‑B‑C” is different from “C‑B‑A”.
Two main concepts are used:
1. Permutations – arrangements of all objects.
2. Combinations – selections of objects without regard to order.
In logical‑reasoning problems we often mix these ideas with constraints (e.g., “two people must sit together”) or with identical items.

2. Key Points

Concept	Symbol	Formula	Example
Permutations of distinct items	(P(n,r)) or (nPr)	(\displaystyle nPr = \frac{n!}{(n-r)!})	5 people in a row: (5P5=5!=120)
Permutations of all distinct items	(n!)	–	4 books on a shelf: (4!=24)
Combinations	(C(n,r)) or (\displaystyle\binom{n}{r})	(\displaystyle \frac{n!}{r!(n-r)!})	3 out of 10 students chosen for a team: (\binom{10}{3}=120)
Arrangements with identical items	Use division by factorials of identicals	(\displaystyle \frac{n!}{n_1!n_2!\dots})	2 A’s, 1 B, 1 C → (\frac{4!}{2!}=12)
Restricted arrangements	Treat restricted objects as a single entity or use inclusion‑exclusion	Varies	“Two people must sit together” → treat them as one, then multiply by 2! for their internal order.
Circular arrangements	((n-1)!) for distinct items	–	6 friends sitting around a round table: (5!=120)
Linear arrangements with gaps	Place “gaps” between items to insert others	–	3 men seated → 4 gaps → place 2 women in gaps.

Important note:

Factorial ((n!)) = (n \times (n-1) \times \dots \times 1).
For small numbers, you can count directly; for larger (n), use the formulas.

3. Worked Example

Problem (Typical SNAP style)

*Six people (A, B, C, D, E, F) are to be seated in a row.

A and B must not sit next to each other.

C must be to the left of D.
How many seating arrangements satisfy both conditions?*

Step‑by‑step Solution

Total arrangements without any restrictions
(6! = 720).
Exclude arrangements where A and B sit together
- Treat A‑B as a single block.
- We now have 5 items: (AB), C, D, E, F.
- Arrangements of these 5 items: (5! = 120).
- Inside the block, A and B can swap: (2!) ways.
- So, (\text{A‑B together} = 5! \times 2! = 240).
Apply the condition “C left of D”
- For any arrangement of the 6 distinct people, exactly half have C left of D (since swapping C and D gives a distinct arrangement).
- Thus, any set of arrangements that is symmetric in C and D can be halved.
Combine the two restrictions
- First, count arrangements where A and B are not together but C can be on either side of D.
  [ \text{Not together} = 720 - 240 = 480 ]
- Now enforce “C left of D” by halving:
  [ \frac{480}{2} = 240 ]

Answer: 240 seating arrangements meet both conditions.

Quick Verification

If we first enforce “C left of D” (half of 720 = 360) and then remove “A‑B together” inside that set:
- Count arrangements with A‑B together and C left of D:
  - Treat A‑B as block → 5 items → (5! = 120).
  - Half of these have C left of D: (120/2 = 60).
- Subtract: (360 - 60 = 300).
- Wait! This gives 300, not 240.
- Why the mismatch? Because the two restrictions are not independent: when A‑B together, the “C left of D” half‑rule may not hold if C or D is part of the block.
- The correct method is the first one (remove A‑B together first, then apply C‑D ordering).

4. Quick Revision Summary

Permutations: order matters → (n!) or (nPr).
Combinations: order doesn’t matter → (\binom{n}{r}).
Identical items: divide by factorials of identical counts.
Restrictions:
- Together: treat as a block.
- Not together: total minus “together”.
- Order constraints: often half the total (swap the two items).