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Simple Combinations (Order Doesn’t Matter)

Introduction

Simple Combinations तब उपयोग होती हैं जब हम items select करते हैं और order matter नहीं करता। Combinations तब use करें जब सिर्फ यह मायने रखता हो कि कौन-कौन से items चुने गए हैं - न कि वे किस क्रम में आते हैं।

यह pattern महत्वपूर्ण है क्योंकि कई real-world selection problems (team बनाना, committee members चुनना, lottery numbers select करना) permutations नहीं बल्कि combinations मांगती हैं।

Pattern: Simple Combinations (Order Doesn’t Matter)

Pattern

अगर n distinct items में से r items चुनने हों और order irrelevant हो, तो nCr का उपयोग करें।

Formula:
nCr = n! / (r! (n - r)!)

Shortcut idea: इसे nPr / r! के रूप में भी compute कर सकते हैं (हर selection के r! possible orders को divide कर देते हैं)।

Step-by-Step Example

Question

6 students में से 3 students की एक team चुननी है। Team कितने तरीकों से चुनी जा सकती है?

Solution

  1. Step 1: दिए गए values पहचानें।

    Total students = n = 6; team size = r = 3. Order matter नहीं करता।

  2. Step 2: Combination formula चुनें।

    Use nCr = n! / (r! (n - r)!).

  3. Step 3: Substitute करें और compute करें।

    6C3 = 6! / (3! × 3!) = (6 × 5 × 4 × 3!)/(3! × 3!) = (6 × 5 × 4) / (3 × 2 × 1) = 120 / 6 = 20

  4. Final Answer:

    Team 20 तरीकों से चुनी जा सकती है।

  5. Quick Check:

    सोचें: चुने गए 3 students की permutations = 6 × 5 × 4 = 120 इन्हें 3! = 6 से divide करें (order हटाने के लिए) → 120 ÷ 6 = 20 ✅

Quick Variations

1. r = 1 → nC1 = n (simple)।

2. r = n → nCn = 1 (सभी चुनने का सिर्फ एक तरीका)।

3. Committees, lottery picks, unordered groups select करने में combinations उपयोग करें; अगर roles (order) assign होते हों तो permutations उपयोग करें।

Trick to Always Use

  • Step 1: पूछें: "क्या order matter करता है?" अगर NO → combinations (nCr)।
  • Step 2: nCr को factorials से या फिर nPr ÷ r! से (छोटे r पर तेज़) compute करें।

Summary

Summary

n में से r items चुनते समय जब order की कोई भूमिका नहीं होती:

  • Use nCr = n! / (r! (n - r)!).
  • Alternatively, permutations को r! से divide करके ordering हटा दें।
  • हमेशा confirm करें कि problem में order चाहिए या नहीं - यही combination vs permutation तय करता है।

Practice

(1/5)
1. From a group of 5 students, how many ways can 2 students be chosen for a project team?
easy
A. 10
B. 8
C. 12
D. 6

Solution

  1. Step 1: Identify what is given.

    n = 5 students; r = 2 to choose; order does not matter.

  2. Step 2: Apply the combination formula.

    nCr = n! / (r! (n - r)!).

  3. Step 3: Substitute and compute.

    5C2 = 5! / (2! × 3!) = (5 × 4) / (2 × 1) = 10.

  4. Final Answer:

    There are 10 ways → Option A.

  5. Quick Check:

    5P2 = 20; divide by 2! = 10 ✅
Hint: Use nC2 = n(n - 1)/2 for quick pair counts.
Common Mistakes: Using permutation (nPr) instead of combination.
2. From 4 applicants, in how many ways can a 2-person panel be chosen (order doesn't matter)?
easy
A. 12
B. 6
C. 4
D. 8

Solution

  1. Step 1: Identify what is given.

    n = 4; r = 2; order does not matter.

  2. Step 2: Apply the combination formula.

    4C2 = 4! / (2! × 2!).

  3. Step 3: Substitute and compute.

    4C2 = (4 × 3) / (2 × 1) = 6.

  4. Final Answer:

    There are 6 ways → Option B.

  5. Quick Check:

    Pairs: (1,2),(1,3),(1,4),(2,3),(2,4),(3,4) → 6 ✅
Hint: For small n, list pairs briefly to verify counts.
Common Mistakes: Confusing order - counting (A,B) and (B,A) as different.
3. From 8 players, how many teams of 4 players can be formed?
easy
A. 60
B. 80
C. 70
D. 90

Solution

  1. Step 1: Identify what is given.

    n = 8; r = 4; order does not matter.

  2. Step 2: Apply the combination formula.

    8C4 = 8! / (4! × 4!).

  3. Step 3: Substitute and compute.

    8C4 = (8 × 7 × 6 × 5) / (4 × 3 × 2 × 1) = 1680 / 24 = 70.

  4. Final Answer:

    There are 70 teams → Option C.

  5. Quick Check:

    8P4 = 1680; divide by 4! = 24 → 1680 ÷ 24 = 70 ✅
Hint: Cancel factorial terms early to simplify calculations.
Common Mistakes: Computing full factorials without cancellation leading to errors.
4. From 7 books, how many ways can you choose 3 to take on holiday (order irrelevant)?
medium
A. 21
B. 25
C. 28
D. 35

Solution

  1. Step 1: Identify what is given.

    n = 7 books; r = 3; order does not matter.

  2. Step 2: Apply the combination formula.

    7C3 = 7! / (3! × 4!).

  3. Step 3: Substitute and compute.

    7C3 = (7 × 6 × 5) / (3 × 2 × 1) = 210 / 6 = 35.

  4. Final Answer:

    There are 35 ways → Option D.

  5. Quick Check:

    7P3 = 7 × 6 × 5 = 210; divide by 3! = 6 → 210 ÷ 6 = 35 ✅
Hint: Use nC3 = n(n - 1)(n - 2) / 6 for quick computation.
Common Mistakes: Treating ordered arrangements as combinations or vice versa.
5. A box contains 10 different chocolates. In how many ways can 3 chocolates be selected?
medium
A. 120
B. 100
C. 90
D. 720

Solution

  1. Step 1: Identify what is given.

    n = 10; r = 3; order does not matter.

  2. Step 2: Apply the combination formula.

    10C3 = 10! / (3! × 7!) = (10 × 9 × 8) / (3 × 2 × 1).

  3. Step 3: Compute.

    (10 × 9 × 8) / 6 = 720 / 6 = 120.

  4. Final Answer:

    There are 120 ways → Option A.

  5. Quick Check:

    10P3 = 720; divide by 3! = 6 → 720 ÷ 6 = 120 ✅
Hint: Compute numerator as n(n - 1)(n - 2) and divide by 6 for r = 3.
Common Mistakes: Using permutation (nPr) or misplacing factorial division.

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