0
0

Probability Using Permutations/Combinations

Introduction

कई probability problems में items को चुनने या arrange करने के तरीकों की संख्या निकालनी होती है। जब order matter नहीं करता, तब combinations (nCr) का उपयोग किया जाता है। जब order matter करता है, तब permutations (nPr) का उपयोग होता है। यह pattern selection, arrangement, committee formation और lottery-style problems के लिए बहुत महत्वपूर्ण है।

Logic simple है - favourable और total outcomes को permutation/combination formulas से count करें, और फिर यह formula apply करें: P(E) = Favourable Outcomes ÷ Total Outcomes.

Pattern: Probability Using Permutations/Combinations

Pattern

पहले तय करें कि order matter करता है या नहीं। Order matter करने पर permutations; order matter न करने पर combinations। फिर P = favourable ÷ total apply करें।

  • Combinations (nCr): जब selection का order matter नहीं करता → nCr = n! / (r!(n-r)!).
  • Permutations (nPr): जब order matter करता है → nPr = n! / (n-r)!

Step-by-Step Example

Question

(i) 52 cards की deck से बिना replacement 2 cards random निकाले जाते हैं। दोनों Aces होने की probability क्या है?
(ii) 5 students में से एक President और एक Secretary चुने जाते हैं। Alice President और Bob Secretary बनने की probability क्या है?

Solution

  1. Part (i): Using Combinations (order matter नहीं करता)

    Step 1: 2 cards चुनने के total ways = 52C2 = 1326.

    Step 2: 2 Aces चुनने के favourable ways = 4C2 = 6.

    Step 3: Probability = 6 ÷ 1326 = 1/221.

  2. Quick Check (Alternative Method)

    Sequentially: (4/52) × (3/51) = 12/2652 = 1/221 ✅

  3. Part (ii): Using Permutations (order matter करता है)

    Step 1: Total arrangements = 5P2 = 5 × 4 = 20.

    Step 2: Favourable arrangement (Alice-President, Bob-Secretary) = 1.

    Step 3: Probability = 1 ÷ 20 = 1/20.

  4. Final Answers:

    (i) दोनों Aces = 1/221
    (ii) Alice-Bob pair = 1/20

  5. Quick Check:

    दोनों results alternative verification से match करते हैं ✅

Quick Variations

1. Teams या committees select करना → combinations use करें।

2. Seats arrange करना या ranks assign करना → permutations use करें।

3. Cards या balls बिना replacement draw करना → आमतौर पर combinations।

4. Multi-step arrangements → ज़रूरत अनुसार nCr और nPr दोनों combine करें।

Trick to Always Use

  • Step 1: पूछें - “क्या order matter करता है?” इससे nCr और nPr में सही चुनाव होगा।
  • Step 2: Factorial formulas से total और favourable outcomes निकालें।
  • Step 3: जब doubt हो (खासकर बिना replacement वाले cases में) sequential probability से verify करें।

Summary

Summary

Probability using Permutations and Combinations pattern में:

  • nCr तब use करें जब order matter नहीं करता।
  • nPr तब use करें जब order matter करता है।
  • Total और favourable outcomes हमेशा सही count करें।
  • P(E) = Favourable / Total apply करें।
  • Replacement या sequence आने पर sequential multiplication से cross-check करें।

Practice

(1/5)
1. Two cards are drawn at random from a pack of 52 cards. What is the probability that both cards are Kings?
easy
A. 1/221
B. 1/325
C. 1/169
D. 1/663

Solution

  1. Step 1: Identify total outcomes

    Total ways to choose 2 cards = 52C2 = 1326.

  2. Step 2: Identify favourable outcomes

    Ways to choose 2 Kings from 4 = 4C2 = 6.

  3. Step 3: Apply formula

    P(both Kings) = 6 ÷ 1326 = 1/221.

  4. Final Answer:

    1/221 → Option A.

  5. Quick Check:

    Sequential check: (4/52) × (3/51) = 12/2652 = 1/221 ✅
Hint: Use combinations for 'without replacement' pairs: favourable = nCr(4,2), total = nCr(52,2).
Common Mistakes: Using replacement logic or forgetting to use combinations for unordered draws.
2. From 6 students, a leader and an assistant leader are to be chosen. What is the probability that Ravi is the leader and Meena is the assistant?
easy
A. 1/15
B. 1/30
C. 1/20
D. 1/10

Solution

  1. Step 1: Determine total possible arrangements

    Total ordered selections = 6P2 = 6 × 5 = 30.

  2. Step 2: Identify favourable arrangement

    Only one ordered arrangement corresponds to Ravi as leader and Meena as assistant → favourable = 1.

  3. Step 3: Apply formula

    P = 1 ÷ 30 = 1/30.

  4. Final Answer:

    1/30 → Option B.

  5. Quick Check:

    One specific ordered pair among 30 possible ordered pairs → 1/30 ✅
Hint: When order matters, use permutations (nP r) for total arrangements.
Common Mistakes: Using combinations instead of permutations when order matters.
3. Out of 8 people, a committee of 3 members is to be formed. What is the probability that a particular person, Ramesh, is included in the committee?
easy
A. 3/8
B. 5/8
C. 1/8
D. 1/4

Solution

  1. Step 1: Total combinations

    Total ways to form a 3-member committee = 8C3 = 56.

  2. Step 2: Favourable combinations (Ramesh included)

    If Ramesh is included, choose remaining 2 from the other 7 → 7C2 = 21.

  3. Step 3: Apply formula

    P(Ramesh included) = 21 ÷ 56 = 3/8.

  4. Final Answer:

    3/8 → Option A.

  5. Quick Check:

    21 favourable committees out of 56 total → 21/56 = 3/8 ✅
Hint: Fix the required person, then choose the remaining members from the rest using nCr.
Common Mistakes: Using 8C2 instead of 7C2 for favourable cases when one member is fixed.
4. From 10 people, a 4-member team is to be chosen. What is the probability that two specific people, A and B, are always included?
medium
A. 1/15
B. 1/6
C. 2/15
D. 1/5

Solution

  1. Step 1: Total teams

    Total possible 4-member teams = 10C4 = 210.

  2. Step 2: Favourable teams (A and B fixed)

    Fix A and B, choose remaining 2 from the other 8 → 8C2 = 28.

  3. Step 3: Apply formula

    P(A and B included) = 28 ÷ 210 = 2/15.

  4. Final Answer:

    2/15 → Option C.

  5. Quick Check:

    28 favourable teams out of 210 total → 28/210 = 2/15 ✅
Hint: Fix required members and use nCr to choose the rest from remaining people.
Common Mistakes: Forgetting to fix A and B and instead choosing them again as part of selection.
5. A box has 5 red, 4 green, and 3 blue balls. If 3 balls are drawn at random without replacement, what is the probability that all are red?
medium
A. 1/24
B. 1/28
C. 1/30
D. 1/22

Solution

  1. Step 1: Identify total outcomes

    Total balls = 5 + 4 + 3 = 12. Total ways to draw 3 = 12C3 = 220.

  2. Step 2: Identify favourable outcomes

    Ways to draw 3 red balls from 5 = 5C3 = 10.

  3. Step 3: Apply formula

    P(all red) = 10 ÷ 220 = 1/22.

  4. Final Answer:

    1/22 → Option D.

  5. Quick Check:

    10 favourable combinations out of 220 total → 10/220 = 1/22 ✅
Hint: Use nCr for both total and favourable counts when drawing without replacement.
Common Mistakes: Treating draws as independent (replacement) and squaring single-draw probabilities.

Mock Test

Ready for a challenge?

Take a 10-minute AI-powered test with 10 questions (Easy-Medium-Hard mix) and get instant SWOT analysis of your performance!

10 Questions
5 Minutes