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Probability Using Permutations/Combinations

Introduction

Many probability problems ask for the number of ways to choose or arrange items. When order does not matter, use combinations (nCr). When order matters, use permutations (nPr). This pattern is essential for questions about selection, arrangement, committee formation, and lottery-style problems.

The logic is simple - count favourable and total outcomes using permutation or combination formulas, and then apply: P(E) = Favourable Outcomes ÷ Total Outcomes.

Pattern: Probability Using Permutations/Combinations

Pattern

Decide whether order matters. Use permutations when order matters; use combinations when order does not matter. Then apply P = favourable ÷ total.

  • Combinations (nCr): Use when the order of selection doesn’t matter → nCr = n! / (r!(n-r)!).
  • Permutations (nPr): Use when order matters → nPr = n! / (n-r)!

Step-by-Step Example

Question

(i) Two cards are drawn at random from a deck of 52 cards without replacement. What is the probability both are Aces?
(ii) From 5 students, a President and a Secretary are to be chosen. What is the probability that Alice becomes President and Bob becomes Secretary?

Solution

  1. Part (i): Using Combinations (order doesn’t matter)

    Step 1: Total ways to choose 2 cards = 52C2 = 1326.

    Step 2: Favourable ways to choose 2 Aces = 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 matters)

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

  5. Quick Check:

    Both results match alternative verification ✅

Quick Variations

1. Selecting teams or committees → Use combinations.

2. Arranging seats or assigning ranks → Use permutations.

3. Drawing cards or balls without replacement → Usually combinations.

4. Multi-step arrangements → Combine nCr and nPr as needed.

Trick to Always Use

  • Step 1: Ask “Does order matter?” to choose between nCr and nPr.
  • Step 2: Compute total and favourable outcomes using factorial formulas.
  • Step 3: Verify by sequential probability when unsure (especially without replacement).

Summary

Summary

In the Probability using Permutations and Combinations pattern:

  • Use nCr for combinations (order doesn’t matter).
  • Use nPr for permutations (order matters).
  • Always compute both total and favourable outcomes correctly.
  • Apply P(E) = Favourable / Total.
  • Cross-check with sequential multiplication when replacement or sequence is involved.

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

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