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Simple Combinations (Order Doesn’t Matter)

Introduction

Simple Combinations help when we select items and the order does not matter. Use combinations when you care only about which items are chosen - not the sequence in which they appear.

This pattern is important because many real-world selection problems (forming teams, choosing committee members, selecting lottery numbers) require combinations rather than permutations.

Pattern: Simple Combinations (Order Doesn’t Matter)

Pattern

If you need to choose r items from n distinct items and order is irrelevant, use nCr.

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

Shortcut idea: Compute as nPr / r! (permutations divided by the r! orders of each selection).

Step-by-Step Example

Question

From 6 students, a team of 3 students is to be chosen. In how many ways can the team be selected?

Solution

  1. Step 1: Identify what is given.

    Total students = n = 6; team size = r = 3. Order does not matter.

  2. Step 2: Choose the combination formula.

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

  3. Step 3: Substitute and compute.

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

  4. Final Answer:

    The team can be selected in 20 different ways.

  5. Quick Check:

    Verify by thinking: number of permutations for 3 chosen (6 × 5 × 4 = 120) divided by 3! = 6 (orders of each team) → 120 ÷ 6 = 20 ✅

Quick Variations

1. Choosing r = 1 → nC1 = n (trivial).

2. Choosing r = n → nCn = 1 (only one way to choose all).

3. Use combinations for selecting committees, lottery picks, or unordered groups; use permutations if roles (order) are assigned within the selection.

Trick to Always Use

  • Step 1: Ask: "Does order matter?" If NO → combinations (nCr).
  • Step 2: Compute nCr either directly with factorials or as nPr ÷ r! for speed when r is small.

Summary

Summary

For selecting r items from n without caring about order:

  • Use nCr = n! / (r! (n - r)!).
  • Equivalently, compute permutations then divide by r! to remove ordering.
  • Always confirm whether the problem requires order - that decides combination vs permutation.

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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