1st Year Math Chapter 7 Exercise 7.4 Notes: Combinations and Complementary Combination (Punjab Board 2026-27)
Complete concept notes, formulas, and solved examples for ICS/FSc Part 1 Mathematics, Punjab Board (PECTAA), Single National Curriculum 2026-27 session.
Exercise 7.4 closes the chapter by introducing combinations — selections of objects where, unlike permutations, the order does not matter — along with the useful complementary combination property.
What Does This Exercise Cover?
This exercise teaches the combination formula for selecting r objects from n distinct objects without regard to order, and shows how selecting r objects is mathematically equivalent to leaving out the remaining (n − r).
Key Concepts
1. What Is a Combination?
A combination is a selection of objects where the order does not matter. Choosing the same group of objects in a different order does not count as a new combination.
2. The Combination Formula
The number of ways to choose r objects from n distinct objects, denoted nCr, is given by nCr = n! / (r!(n − r)!).
3. Complementary Combination
Choosing r objects to include is equivalent to choosing (n − r) objects to leave out, which gives the identity nCr = nC(n−r).
4. Combination vs Permutation
The key distinguishing question is whether order matters: if yes, use a permutation (nPr); if no, use a combination (nCr).
Solved Examples
Example 1: Evaluate 6C2.
6C2 = 6! / (2! × 4!) = 720 / (2 × 24).
Answer: 15
Example 2: In how many ways can a committee of 3 be selected from 8 people?
8C3 = 8! / (3! × 5!) = 40320 / (6 × 120).
Answer: 56 ways
Example 3: Verify that 10C3 = 10C7 using the complementary combination property.
10C3 = 10! / (3! × 7!) = 120.
10C7 = 10! / (7! × 3!) = 120.
Answer: Both equal 120, confirming 10C3 = 10C7
Sample MCQs
1. A combination is a selection where:
a) Order matters b) Order does not matter c) Repetition is required d) All objects must be used
Answer: b) Order does not matter
2. The formula for nCr is:
a) n! / (n − r)! b) n! / (r!(n − r)!) c) n! × r! d) n! − r!
Answer: b) n! / (r!(n − r)!)
3. The value of 5C2 is:
a) 10 b) 15 c) 20 d) 25
Answer: a) 10
4. The complementary combination property states:
a) nCr = nC(n−r) b) nCr = nPr c) nCr = n!/r! d) nCr = 1
Answer: a) nCr = nC(n−r)
5. The number of ways to choose a committee of 3 from 8 people is:
a) 336 b) 56 c) 24 d) 8
Answer: b) 56
Important Short Questions
- Define a combination and explain how it differs from a permutation.
- State the formula for nCr.
- Evaluate 7C3.
- State the complementary combination property in your own words.
- In how many ways can a team of 2 be chosen from 6 players?
Important Long Questions
- In how many ways can a committee of 4 be selected from 10 people?
- Evaluate 9C4, and verify it equals 9C5 using the complementary combination property.
- A box contains 5 red balls and 3 blue balls. In how many ways can 3 balls be selected so that all 3 are red?
- Explain, with an example, why order matters in a permutation but not in a combination.
How to Approach This Exercise Effectively
- Ask ‘does the arrangement matter, or just the group chosen?’ — this single question separates permutation problems from combination problems.
- Use the complementary combination property (nCr = nC(n−r)) to simplify calculations whenever r is closer to n than to 0.
- Practice canceling factorials in nCr calculations before multiplying everything out — it keeps numbers manageable.
- For committee or team-selection problems, remind yourself that swapping two chosen people around doesn’t create a new committee — that’s why it’s a combination, not a permutation.
- When a problem restricts the type of item chosen (like ‘all red balls’), apply the combination formula only to the relevant subgroup, not the whole set.
FAQs
Q: What is the key difference between a permutation and a combination?
A: A permutation counts arrangements where order matters, while a combination counts selections where only the group chosen matters, not their order.
Q: Why does nCr equal nC(n−r)?
A: Choosing which r objects to include is exactly equivalent to choosing which (n−r) objects to exclude, so both selections are counted by the same number of ways.
Q: How is nCr related to nPr?
A: nPr counts ordered arrangements, while nCr counts unordered selections; specifically, nPr = nCr × r!, since each combination of r objects can be arranged in r! different orders.
Q: Can nCr ever be greater than nPr for the same n and r?
A: No — since nPr = nCr × r! and r! is at least 1, nPr is always greater than or equal to nCr for the same n and r (they’re equal only when r = 0 or r = 1).
Notes prepared for Punjab Board (PECTAA) 1st Year Mathematics, 2026-27 session, Chapter 7: Permutations and Combinations, Exercise 7.4.