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Conditional Permutations (With/Without Restrictions)

Introduction

कई permutation problems में extra rules होते हैं - जैसे कि लोग एक साथ बैठना चाहिए, कुछ items साथ नहीं आ सकते, positions fixed हों, या alternating patterns हों। ये conditional permutations स्पष्ट case-work में टूटकर हल करने पड़ते हैं और हर case में बुनियादी permutation नियम लागू करने होते हैं।

यह pattern महत्वपूर्ण है क्योंकि असली exam questions अक्सर restrictions डालकर counting को जटिल बनाते हैं; systematic case-work और reductions सीखने से गलतियाँ कम होती हैं।

Pattern: Conditional Permutations (With/Without Restrictions)

Pattern

मुख्य विचार: restriction को किसी समतुल्य counting step में बदल दें (किसी block को एक item मान लेना, forbidden arrangements को total से subtract करना, positions को fix करना, या independent choices का multiplication)।

आम approaches:

  • Together: जो items साथ होना चाहिए उन्हें एक single block मानें, block के अंदर के arrangements गिनें, फिर block को बाकी items के साथ arrange करें।
  • Not together: पहले total arrangements निकालें और उन arrangements को subtract करें जहाँ items साथ हैं (Total - Together)।
  • Fixed positions: constrained items को पहले place करें (anchor), फिर बाकी slots permute करें।
  • Alternating arrangement: पहले एक type (जैसे women) को चुने हुए slots में रखें, फिर दूसरी type को बाकी slots में permute करें।

Step-by-Step Example

Question

पाँच लोग A, B, C, D, E एक row में बैठे हैं। अगर A और B को साथ बैठना अनिवार्य हो तो कितने arrangements संभव हैं?

Solution

  1. Step 1: restriction समझें।

    A और B को adjacent होना चाहिए → AB को एक single block मानें (इसे X कहें)। अब X, C, D, E को arrange करें।

  2. Step 2: blocks arrange करें।

    X, C, D, E को arrange करने के तरीके = 4! = 24.

  3. Step 3: block के अंदर की internal arrangements गिनें।

    X के अंदर A और B दो गो तरीकों से होंगे: AB या BA → 2! = 2.

  4. Step 4: स्वतंत्र counts को multiply करें।

    कुल arrangements = blocks के arrangements × internal arrangements = 4! × 2 = 24 × 2 = 48.

  5. Final Answer:

    ऐसे 48 valid arrangements हैं जहाँ A और B साथ बैठते हैं।

  6. Quick Check:

    अगर A और B free होते तो कुल 5! = 120 होते; together-case इससे छोटा होना चाहिए: 48 < 120 ✅

Quick Variations

1. “Never together”: कुल - together. उदाहरण: total = 5! = 120; together = 48 → never together = 72।

2. Three together: तीन को एक block मानें; internal 3! arrangements से multiply करें।

3. Fixed seats: अगर किसी व्यक्ति को कोई particular seat चाहिए, तो उसे पहले fix करें और बाकी (n - 1)! तरीके से permute करें।

4. Alternate men & women: छोटी group को पहले alternate slots में रखें, फिर दूसरी group को बाकी में arrange करें - उदाहरण: 3 men और 3 women के लिए row या table में placements इस तरह गिने जाते हैं।

Trick to Always Use

  • Step 1 → restriction को translate करें: block बनाना है, exclusion करना है, fixed slot है या alternating slots चाहिए।
  • Step 2 → transformed problem के arrangements गिनें (factorials या permutations उपयोग करें)।
  • Step 3 → अंदर के arrangements (blocks के internal order) और उस block/people के लिए position choices से multiply करें।
  • Step 4 → quick sanity check करें: परिणाम unconstrained total (n!) से ≤ होना चाहिए।

Summary

Summary

Conditional permutations को नीचे दिए ट्रांसफ़ॉर्म्स में से किसी एक से सरल counting में बदला जा सकता है:

  • Together: grouped items को block मानें → arrange blocks × internal permutations।
  • Not together: Total - Together।
  • Fixed positions: constrained items पहले place करें, फिर बाकी permute करें।
  • Alternating: एक group को चुने हुए slots में रखें, फिर दूसरी group को बाकी में arrange करें।

हमेशा छोटा case-work स्पष्ट लिखें और unconstrained total के साथ quick check करें।

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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