Maths Extension 1 Year 11 - Module 4 - Lesson 14
Combinatorial Identities & Proofs
Name
Date
Class
1. Lesson focus
Combinatorial identities and proofs. Use symmetry and Pascal's identity to transform combinations; binomial coefficient sums can be read from (1+1)^n.
Success criteria
- Apply the lesson method accurately in a routine example.
- Show each algebraic or domain step and check the result.
- Explain why the method and restrictions fit the question.
2. Worked practice
2. Evaluate Σ(r=0 to 5) (−1)^r 5Cr.
3. Show 8C3=7C2+7C3 numerically.
3. Challenge
Prove nC0+nC1+…+nCn=2ⁿ using the binomial theorem.
4. Multiple choice
For n>0, the alternating sum Σ(r=0 to n)(−1)^r nCr equals what?
A. 2ⁿ
B. 1
C. 0
D. n!
Worked answer key
Compare each step with your working. Keep stated intervals, domains and restrictions in your final answer.
1. Evaluate Σ(r=0 to 6) 6Cr.
By the binomial theorem, the sum is (1+1)⁶=64.
2. Evaluate Σ(r=0 to 5) (−1)^r 5Cr.
By (1−1)⁵=0, the alternating sum is 0.
3. Show 8C3=7C2+7C3 numerically.
8C3=56; 7C2=21 and 7C3=35, so 21+35=56.
Challenge. Prove nC0+nC1+…+nCn=2ⁿ using the binomial theorem.
Expand (1+1)ⁿ=Σ nCr·1ⁿ⁻ʳ·1ʳ; hence the sum equals 2ⁿ.
Multiple choice. For n>0, the alternating sum Σ(r=0 to n)(−1)^r nCr equals what?
Correct answer: C. The binomial theorem gives (1−1)ⁿ=0 for positive n.
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