Topic 6 of 9 · Form 4
Permutations, combinations and series
ⁿPᵣ = n!/(n − r)! (order matters); ⁿCᵣ = n!/(r!(n − r)!) (order does not matter); arrangements with restrictions (together, apart, ends).
Notes for this topic 1 file, in order
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Permutations and combinations
ⁿPᵣ = n!/(n − r)! (order matters); ⁿCᵣ = n!/(r!(n − r)!) (order does not matter); arrangements with restrictions (together, apart, ends).
Worked example: How many 4-letter arrangements can be made from the letters of NUMBERS (all different)?
⁷P₄ = 7 × 6 × 5 × 4 = 840
Worked example: A committee of 5 is chosen from 6 men and 4 women. How many include exactly 2 women?
⁴C₂ × ⁶C₃ = 6 × 20 = 120
Binomial expansion
(a + b)ⁿ for positive integer n using ⁿCᵣ; finding a specific term or coefficient.
Worked example: Find the term in x³ in (2 + x)⁶.
⁶C₃ × 2³ × x³ = 20 × 8x³ = 160x³
Arithmetic and geometric progressions
AP: uₙ = a + (n − 1)d; Sₙ = (n/2)[2a + (n − 1)d]. GP: uₙ = arⁿ⁻¹; Sₙ = a(1 − rⁿ)/(1 − r); S∞ = a/(1 − r) for |r| < 1.
Worked example: A GP has first term 54 and common ratio ⅓. Find the sum to infinity.
54 ÷ (1 − ⅓) = 81
Check yourself
- Evaluate ⁸C₃.
- Find the sum of the first 20 terms of 3 + 7 + 11 + …
- In how many ways can 5 people sit in a row if two particular people must sit together?
Show answers
- 56
- 10(6 + 76) = 820
- 4! × 2! = 48
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