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Conditional Permutations (With/Without Restrictions)

Introduction

Many permutation problems include extra rules - people who must sit together, items that can’t be adjacent, fixed positions, or alternating patterns. These conditional permutations require breaking the problem into clear cases and applying basic permutation rules within each case.

This pattern is important because real exam questions often add restrictions to make counting non-trivial; learning systematic case-work and reductions prevents mistakes.

Pattern: Conditional Permutations (With/Without Restrictions)

Pattern

Key idea: convert the restriction into an equivalent counting step (treat a block as one item, subtract forbidden arrangements from total, fix positions, or use multiplication of independent choices).

Common approaches:

  • Together: Treat the required-together items as a single block, count internal arrangements, then multiply by arrangements of the block with others.
  • Not together: Count total arrangements and subtract arrangements where the items are together (Total - Together).
  • Fixed positions: Place constrained items first (anchor), then permute remaining slots.
  • Alternating arrangement: Place one type (e.g., women) in chosen slots, then permute the other type into remaining slots.

Step-by-Step Example

Question

Five people A, B, C, D, E sit in a row. How many arrangements are possible if A and B must sit together?

Solution

  1. Step 1: Understand the restriction.

    A and B must be adjacent → treat AB as a single block (call it X). Now count arrangements of X, C, D, E.

  2. Step 2: Arrange the blocks.

    Number of ways to arrange X, C, D, E = 4! = 24.

  3. Step 3: Count internal arrangements of the block.

    Inside block X, A and B can be ordered as AB or BA → 2! = 2.

  4. Step 4: Multiply independent counts.

    Total arrangements = arrangements of blocks × internal arrangements = 4! × 2 = 24 × 2 = 48.

  5. Final Answer:

    48 valid arrangements where A and B sit together.

  6. Quick Check:

    If A and B were free → 5! = 120. Together-case should be smaller: 48 < 120 ✅

Quick Variations

1. “Never together”: Total - Together. Example: total 5! = 120; together = 48 → never together = 72.

2. Three together: Treat three as a single block; multiply by internal 3! arrangements.

3. Fixed seats: If a person must sit in a particular seat, fix them first and permute remaining (n - 1)! ways.

4. Alternate men & women: Place the smaller group first into alternate slots, then permute the other group into remaining slots. Example: for 3 men and 3 women around a table (or row), count placements accordingly.

Trick to Always Use

  • Step 1 → Translate the restriction into an operation: block, exclusion, fixed slot, or alternating slots.
  • Step 2 → Count arrangements of the transformed problem (use factorials or permutations).
  • Step 3 → Multiply by internal arrangements (order inside blocks) and by choices for which positions the block/people occupy.
  • Step 4 → Quick sanity check: result ≤ total unconstrained arrangements (n!).

Summary

Summary

Conditional permutations reduce to simpler counting after one of these transforms:

  • Together: Treat grouped items as a block → arrange blocks × internal permutations.
  • Not together: Use Total - Together.
  • Fixed positions: Place constrained items first, then permute remaining.
  • Alternating: Place one group into chosen slots, then arrange the other group into remaining slots.

Always show the small case-work explicitly and perform a quick check against the unconstrained total.

Practice

(1/5)
1. Six friends A, B, C, D, E, and F sit in a row. In how many arrangements are A and B always together?
easy
A. 240
B. 360
C. 480
D. 720

Solution

  1. Step 1: Combine A and B into one block.

    Treat A and B as a single unit (X). Now we have X, C, D, E, F → 5 items in total.

  2. Step 2: Arrange the blocks.

    Number of ways to arrange 5 items = 5! = 120.

  3. Step 3: Arrange A and B within their block.

    A and B can be arranged internally in 2! = 2 ways.

  4. Step 4: Multiply both results.

    Total arrangements = 5! × 2 = 120 × 2 = 240.

  5. Final Answer:

    There are 240 valid arrangements → Option A.

  6. Quick Check:

    Without restriction = 6! = 720; together-case should be smaller: 240 < 720 ✅
Hint: Treat the pair as one block, compute factorial of reduced count, and multiply by 2! for internal order.
Common Mistakes: Using 6! directly or forgetting internal swapping of A and B.
2. Four people A, B, C, D sit in a row. In how many arrangements are A and B not sitting together?
easy
A. 6
B. 12
C. 18
D. 24

Solution

  1. Step 1: Count total arrangements without restriction.

    Total = 4! = 24.

  2. Step 2: Count arrangements where A and B sit together.

    Treat AB as a block → items: (AB), C, D → 3! arrangements = 6; internal AB orders = 2 → together = 6 × 2 = 12.

  3. Step 3: Subtract to get 'not together'.

    Not together = Total - Together = 24 - 12 = 12.

  4. Final Answer:

    12 → Option B.

  5. Quick Check:

    Not together + together = 12 + 12 = 24 (matches total) ✅
Hint: Use Total - Together for 'not together' problems.
Common Mistakes: Counting only block arrangements and forgetting internal orders when computing 'together'.
3. Five people sit around a round table. In how many distinct circular arrangements do A and B sit together? (rotations considered identical, reflections distinct)
easy
A. 12
B. 24
C. 48
D. 6

Solution

  1. Step 1: Convert to circular-block problem.

    Treat A and B as a single block. Now objects around circle: (AB), C, D, E → 4 items around a circle.

  2. Step 2: Use circular permutation for blocks.

    Number of circular arrangements of 4 distinct items (rotations identical) = (4 - 1)! = 3! = 6.

  3. Step 3: Multiply by internal orders of block.

    AB can be AB or BA → 2 orders. Total = 6 × 2 = 12.

  4. Final Answer:

    12 → Option A.

  5. Quick Check:

    Anchor one element of the 4-block circle, arrange remaining 3 → 3! = 6; times 2 internal orders = 12 ✅
Hint: For circular 'together' problems, treat block as one then use (m - 1)! for m blocks, and multiply by internal orders.
Common Mistakes: Using linear permutations (m!) instead of circular ((m - 1)!).
4. Seven people are seated in a row. If three specific people must sit together, how many arrangements are possible?
medium
A. 3,600
B. 504
C. 720
D. 1,260

Solution

  1. Step 1: Group the three people as one block.

    Treat the three-person block as a single item. Now total items = block + 4 other people = 5 items.

  2. Step 2: Arrange the blocks in the row.

    Number of ways to arrange 5 items = 5! = 120.

  3. Step 3: Count internal arrangements inside the block.

    The three people can be ordered in 3! = 6 ways.

  4. Step 4: Multiply counts.

    Total arrangements = 5! × 3! = 120 × 6 = 720.

  5. Final Answer:

    720 → Option C.

  6. Quick Check:

    Result ≤ 7! = 5,040 and seems reasonable: 720 < 5,040 ✅
Hint: Treat required-together group as one block, arrange blocks, then multiply by internal permutations.
Common Mistakes: Forgetting internal permutations of the grouped people (3!).
5. Three men and three women are to be seated in a row of six chairs so that sexes alternate. How many arrangements are possible?
medium
A. 36
B. 18
C. 48
D. 72

Solution

  1. Step 1: Identify alternate-slot patterns.

    For a row of six, sexes can alternate in two patterns: M W M W M W or W M W M W M → 2 choices of pattern.

  2. Step 2: Permute men and women within their slots.

    Men can be arranged in 3! = 6 ways; women in 3! = 6 ways.

  3. Step 3: Multiply all choices.

    Total = 2 × 3! × 3! = 2 × 6 × 6 = 72.

  4. Final Answer:

    72 → Option D.

  5. Quick Check:

    Alternate arrangements must be fewer than 6! = 720; 72 is reasonable ✅
Hint: Count slot-pattern choices (usually 2) then multiply by permutations of each group (3! × 3!).
Common Mistakes: Using combinations (ignoring order) instead of permutations for arranging people in slots.

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