The Gap Method & Permutations under Restrictions (Linear Arrangements)
Master the "Tie & Arrange" method for "together" cases and the powerful "Gap Method" for "never together" cases with examples.
Why These Methods Matter for JEE
Permutations with restrictions appear in almost every JEE paper. These 2 methods cover 90% of restriction-based permutation problems:
- Tie & Arrange Method - When specific objects must be together
- Gap Method - When specific objects must never be together
- Mixed Restrictions - Combination of both methods
- JEE Favorite - Appears in both Mains and Advanced
Tie & Arrange Method (Objects MUST be together)
When certain objects must always remain together in the arrangement.
Step-by-Step Approach:
- Tie the objects that must be together as a single unit
- Arrange all units (including the tied unit)
- Arrange internally the objects within the tied unit
- Multiply the results from steps 2 and 3
Example 1: A and B must be together
Arrange 5 people: A, B, C, D, E in a row such that A and B are always together.
Step 1: Tie A and B together as 1 unit: [AB]
Step 2: We now have 4 units to arrange: [AB], C, D, E
• Number of arrangements = $4! = 24$
Step 3: A and B can be arranged internally in 2 ways: AB or BA
Step 4: Total arrangements = $4! × 2! = 24 × 2 = 48$
Example 2: Three specific together
Arrange 6 books on a shelf such that 3 particular books are always together.
Step 1: Tie the 3 books as 1 unit: [XYZ]
Step 2: We now have 4 units: [XYZ] + 3 other books
• Number of arrangements = $4! = 24$
Step 3: The 3 books can be arranged internally in $3! = 6$ ways
Step 4: Total arrangements = $4! × 3! = 24 × 6 = 144$
Gap Method (Objects MUST NOT be together)
When certain objects must never be together in the arrangement.
Step-by-Step Approach:
- Arrange the objects without restrictions first
- Identify gaps between these objects
- Place the restricted objects in these gaps
- Multiply the arrangements
Example 1: A and B must not be together
Arrange 5 people: A, B, C, D, E in a row such that A and B are never together.
Step 1: Total arrangements without restrictions = $5! = 120$
Step 2: Arrangements where A and B are together = $4! × 2! = 48$ (from previous method)
Step 3: Arrangements where A and B are never together = Total - Together
• $120 - 48 = 72$
Alternative Gap Method:
Step 1: Arrange C, D, E first: $3! = 6$ ways
Step 2: These create 4 gaps where A and B can be placed:
Step 3: Choose 2 gaps out of 4 for A and B: $\binom{4}{2} = 6$ ways
Step 4: Arrange A and B in these 2 gaps: $2! = 2$ ways
Step 5: Total = $3! × \binom{4}{2} × 2! = 6 × 6 × 2 = 72$
Example 2: No two vowels together
Arrange letters of "OBJECT" such that no two vowels are together.
Step 1: Vowels: O, E; Consonants: B, J, C, T
Step 2: Arrange consonants: $4! = 24$ ways
B _ J _ C _ T _ (5 gaps created)
Step 3: Choose 2 gaps out of 5 for vowels: $\binom{5}{2} = 10$ ways
Step 4: Arrange vowels: $2! = 2$ ways
Step 5: Total = $4! × \binom{5}{2} × 2! = 24 × 10 × 2 = 480$
🚀 Problem-Solving Strategies
For "Together" Cases:
- Always tie the restricted objects first
- Count the tied unit as one object
- Don't forget internal arrangement of tied objects
- Multiply all arrangement counts
For "Never Together" Cases:
- Use Gap Method for direct approach
- Or use: Total - Together
- Count gaps carefully (n objects create n+1 gaps)
- Remember to arrange restricted objects among themselves
Mixed Restrictions Problems
Combining both methods for complex restrictions.
Example: A&B together, C&D not together
Arrange 6 people A, B, C, D, E, F such that A and B are together, but C and D are never together.
Step 1: Tie A and B together as [AB] unit
Step 2: We have 5 units: [AB], C, D, E, F
Step 3: Total arrangements of 5 units = $5! = 120$
Step 4: Internal arrangement of A and B = $2! = 2$
Step 5: So far: $120 × 2 = 240$ arrangements with A and B together
Step 6: From these, subtract cases where C and D are together
Step 7: With A and B tied, and C and D tied: 4 units = $4! = 24$
Step 8: Internal arrangements: A,B = 2!, C,D = 2! → $24 × 2 × 2 = 96$
Step 9: Final answer = $240 - 96 = 144$
📝 Quick Self-Test
Try these JEE-level problems to test your understanding:
1. Arrange 7 people such that 2 particular are always together and 3 particular are never together.
2. How many words can be formed from "ENGINEER" with all E's together?
3. Arrange 5 boys and 3 girls such that no two girls sit together.
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