Unit 1 Review — Specialist Mathematics

This review spans all three topics in Unit 1 of Specialist Mathematics: Combinatorics (counting principles, permutations, combinations, Pascal’s triangle, and the binomial theorem), Mathematical Induction (summation and divisibility proofs), and Vectors in the Plane (operations, magnitude, unit vectors, dot product, angles, and vector equations of lines). Questions are exam-style and increase in difficulty. Click each question to reveal the full worked solution.

Mixed Review Questions

Fluency
Q1 — Multiplication Principle

A restaurant offers 4 entrees, 6 mains, and 3 desserts. How many different three-course meals are possible?

By the multiplication principle, each stage of the meal is chosen independently:

Total meals = 4 × 6 × 3 = 72
Fluency
Q2 — Permutations

Calculate: (a) P(8, 3) (b) The number of ways 5 people can be arranged in a row.

(a) P(8, 3) = 8! / (8−3)! = 8! / 5! = 8 × 7 × 6 = 336

(b) P(5, 5) = 5! = 5 × 4 × 3 × 2 × 1 = 120
Fluency
Q3 — Combinations

Calculate: (a) C(10, 4) (b) The number of ways to choose a committee of 3 from 9 people.

(a) C(10, 4) = 10! / (4! × 6!) = (10 × 9 × 8 × 7) / (4 × 3 × 2 × 1) = 5040 / 24 = 210

(b) C(9, 3) = 9! / (3! × 6!) = (9 × 8 × 7) / 6 = 504 / 6 = 84
Fluency
Q4 — Pascal’s Triangle and Binomial Coefficients

Use Pascal’s triangle to find: (a) C(6, 2) (b) the value of C(7, 3) + C(7, 4).

(a) In row 6 of Pascal’s triangle (n=6), the entry at position r=2 is:

C(6, 2) = 6! / (2! × 4!) = 15. Row 6: 1, 6, 15, 20, 15, 6, 1.

(b) By Pascal’s identity: C(n, r) + C(n, r+1) = C(n+1, r+1)

C(7, 3) + C(7, 4) = C(8, 4) = 8!/(4!4!) = 70. So the answer is 70.
Understanding
Q5 — Binomial Theorem

Use the binomial theorem to expand (2x − 3)⁴, writing all terms.

(a + b)⁴ = C(4,0)a⁴ + C(4,1)a³b + C(4,2)a²b² + C(4,3)ab³ + C(4,4)b⁴

With a = 2x and b = −3:

= 1(16x⁴) + 4(8x³)(−3) + 6(4x²)(9) + 4(2x)(−27) + 1(81)

= 16x⁴ − 96x³ + 216x² − 216x + 81

(2x − 3)⁴ = 16x⁴ − 96x³ + 216x² − 216x + 81
Understanding
Q6 — Counting with Restrictions

In how many ways can 4 boys and 3 girls sit in a row if all the girls must sit together?

Treat the 3 girls as a single block. Then we arrange 4 boys + 1 block = 5 units in a row.

Arrangements of 5 units = 5! = 120

Within the block, the 3 girls can be arranged in 3! = 6 ways.

Total = 5! × 3! = 120 × 6 = 720
Fluency
Q7 — Induction Base Case and Structure

Write out the complete structure (base case, inductive hypothesis, and what to prove in the inductive step) for proving ∑_k=1ⁿ k³ = [n(n+1)/2]² by induction.

Base case (n = 1):

LHS = 1³ = 1. RHS = [1(2)/2]² = 1² = 1. ✓

Inductive hypothesis: Assume that for some k ≥ 1,

1³ + 2³ + … + k³ = [k(k+1)/2]².

Inductive step: Show that

1³ + 2³ + … + k³ + (k+1)³ = [(k+1)(k+2)/2]².
Understanding
Q8 — Divisibility Proof by Induction

Prove by mathematical induction that 5ⁿ − 1 is divisible by 4 for all n ≥ 1.

Base case (n = 1):

5¹ − 1 = 4 = 4 × 1. Divisible by 4. ✓

Inductive hypothesis: Assume 5^k − 1 = 4m for some integer m.

Inductive step: Show 5^k+1 − 1 is divisible by 4.

5^k+1 − 1 = 5 × 5^k − 1 = 5 × 5^k − 5 + 4 = 5(5^k − 1) + 4

= 5(4m) + 4 = 4(5m + 1)

This is divisible by 4. ✓

Conclusion: By induction, 4 | (5ⁿ − 1) for all n ≥ 1.
Problem Solving
Q9 — Summation Proof by Induction

Prove by induction that ∑_k=1ⁿ k(k+1) = n(n+1)(n+2)/3 for all positive integers n.

Base case (n = 1):

LHS = 1 × 2 = 2. RHS = 1(2)(3)/3 = 2. ✓

Inductive hypothesis: Assume ∑_j=1^k j(j+1) = k(k+1)(k+2)/3.

Inductive step: Show the result holds for n = k+1.

LHS = k(k+1)(k+2)/3 + (k+1)(k+2)

= (k+1)(k+2)[k/3 + 1]

= (k+1)(k+2)(k+3)/3 = RHS ✓

Conclusion: By induction, the formula holds for all n ∈ ℤ⁺.
Fluency
Q10 — Vector Operations and Magnitude

Given a = 3i − 4j and b = −i + 2j, find: (a) a + 2b (b) |a| (c) â (unit vector in the direction of a)

(a) a + 2b = (3, −4) + 2(−1, 2) = (3−2, −4+4) = (1, 0)

(b) |a| = √(3² + (−4)²) = √(9+16) = √25 = 5

(c) â = a/|a| = (3/5, −4/5) = (3/5)i − (4/5)j
Fluency
Q11 — Dot Product

For u = (4, −3) and v = (2, 5), find: (a) u · v (b) the angle between u and v (to the nearest degree).

(a) u · v = 4(2) + (−3)(5) = 8 − 15 = −7

(b) |u| = √(16+9) = 5, |v| = √(4+25) = √29

cosθ = (u · v) / (|u| |v|) = −7 / (5√29) ≈ −0.2600

θ = arccos(−0.2600) ≈ 105°
Understanding
Q12 — Perpendicularity and Dot Product

Find the value of t such that the vectors p = (t, 5) and q = (3, −t) are perpendicular.

Two vectors are perpendicular when their dot product is zero.

p · q = 0

(t)(3) + (5)(−t) = 0

3t − 5t = 0

−2t = 0

t = 0

When t = 0, p = (0, 5) and q = (3, 0), which are clearly perpendicular (parallel to the axes).
Understanding
Q13 — Vector Equation of a Line

Write the vector equation of the line passing through A(2, 5) with direction vector d = (3, −1). Express it also in Cartesian form.

Vector form:

r = (2, 5) + t(3, −1), t ∈ ℝ

or in component form: x = 2 + 3t, y = 5 − t.

Cartesian form: Eliminate t.

From y: t = 5 − y. Substitute into x: x = 2 + 3(5−y) = 17 − 3y

x + 3y = 17, or equivalently y = (17−x)/3 = −(1/3)x + 17/3

Cartesian equation: x + 3y = 17
Understanding
Q14 — Scalar Resolute (Projection)

Find the scalar resolute of a = (6, 2) in the direction of b = (3, 4). Then find the vector projection of a onto b.

|b| = √(9 + 16) = 5. b̂ = (3/5, 4/5).

Scalar resolute: a · b̂ = (6)(3/5) + (2)(4/5) = 18/5 + 8/5 = 26/5

Vector projection:

proj_b a = (a · b / |b|²) b

a · b = 6(3) + 2(4) = 26, |b|² = 25

= (26/25)(3, 4) = (78/25, 104/25)
Problem Solving
Q15 — Intersection of Two Lines

Line ℓ₁: r = (1, 2) + s(2, 1) and Line ℓ₂: r = (3, 0) + t(−1, 3). Find the point of intersection, or determine if the lines are parallel.

Write the parametric equations:

ℓ₁: x = 1 + 2s, y = 2 + s

ℓ₂: x = 3 − t, y = 3t

Direction vectors (2, 1) and (−1, 3) are not parallel (not scalar multiples). So lines are not parallel and must intersect (in the plane).

Set equal: 1 + 2s = 3 − t → 2s + t = 2 … (i)

2 + s = 3t → s − 3t = −2 … (ii)

From (ii): s = 3t − 2. Substitute into (i): 2(3t−2) + t = 2 → 7t = 6 → t = 6/7.

s = 3(6/7) − 2 = 18/7 − 14/7 = 4/7.

Point of intersection using ℓ₁: x = 1 + 2(4/7) = 1 + 8/7 = 15/7, y = 2 + 4/7 = 18/7.

Intersection: (15/7, 18/7)
Problem Solving
Q16 — Specific Term in a Binomial Expansion

Find the term independent of x in the expansion of (x² − 1/x)⁹.

The general term (r+1-th term) of (a + b)⁹ is C(9, r) a^9−r b^r.

With a = x² and b = −1/x = −x⁻¹:

T_r+1 = C(9, r) (x²)^9−r (−x⁻¹)^r

= C(9, r) (−1)^r x^2(9−r) x^−r

= C(9, r) (−1)^r x^18−2r−r

= C(9, r) (−1)^r x^18−3r

For the term independent of x: 18 − 3r = 0 → r = 6.

T₇ = C(9, 6) (−1)⁶ x⁰ = C(9, 3) × 1 = 84

The constant term is 84.
Problem Solving
Q17 — Vector Geometry Proof

Points A, B, C have position vectors a, b, c. Let M be the midpoint of AB. Prove that CM = (1/2)(a + b) − c.

M is the midpoint of AB, so the position vector of M is:

m = (a + b)/2

The vector from C to M is:

CM = m − c = (a + b)/2 − c

= (1/2)(a + b) − c ✓

This is the position vector of M (from the origin) minus the position vector of C, giving the vector from C to M.
Problem Solving
Q18 — Closest Point on a Line to the Origin

Find the point on the line r = (4, 1) + t(3, −5) that is closest to the origin. What is the minimum distance?

A general point on the line is P = (4 + 3t, 1 − 5t).

The vector OP = (4+3t, 1−5t) must be perpendicular to the direction vector (3, −5):

OP · (3, −5) = 0

3(4 + 3t) + (−5)(1 − 5t) = 0

12 + 9t − 5 + 25t = 0

34t + 7 = 0 → t = −7/34

Closest point: x = 4 + 3(−7/34) = 4 − 21/34 = (136−21)/34 = 115/34

y = 1 − 5(−7/34) = 1 + 35/34 = 69/34

Closest point: (115/34, 69/34)

Distance = √((115/34)² + (69/34)²) = (1/34)√(115² + 69²) = (1/34)√(13225 + 4761) = (1/34)√17986 ≈ 3.95 units
Problem Solving
Q19 — Combinatorics Problem Solving

A class has 12 students: 7 boys and 5 girls. In how many ways can a team of 4 be chosen if the team must have at least 2 girls?

Count teams with exactly 2 girls, exactly 3 girls, and exactly 4 girls:

Exactly 2 girls: C(5,2) × C(7,2) = 10 × 21 = 210

Exactly 3 girls: C(5,3) × C(7,1) = 10 × 7 = 70

Exactly 4 girls: C(5,4) × C(7,0) = 5 × 1 = 5

Total = 210 + 70 + 5 = 285
Problem Solving
Q20 — Induction with Inequality

Prove by induction that 2ⁿ ≥ n + 1 for all n ≥ 1.

Base case (n = 1):

2¹ = 2 and 1 + 1 = 2. 2 ≥ 2. ✓

Inductive hypothesis: Assume 2^k ≥ k + 1 for some k ≥ 1.

Inductive step: Show 2^k+1 ≥ k + 2.

2^k+1 = 2 × 2^k ≥ 2(k+1) = 2k + 2

Since k ≥ 1, we have 2k + 2 ≥ k + 2.

Therefore 2^k+1 ≥ k + 2. ✓

Conclusion: By induction, 2ⁿ ≥ n + 1 for all n ≥ 1.

Unit 1 Review — Specialist Mathematics

Mixed Review Questions

Q1 — Multiplication Principle

Q2 — Permutations

Q3 — Combinations

Q4 — Pascal’s Triangle and Binomial Coefficients

Q5 — Binomial Theorem

Q6 — Counting with Restrictions

Q7 — Induction Base Case and Structure

Q8 — Divisibility Proof by Induction

Q9 — Summation Proof by Induction

Q10 — Vector Operations and Magnitude

Q11 — Dot Product

Q12 — Perpendicularity and Dot Product

Q13 — Vector Equation of a Line

Q14 — Scalar Resolute (Projection)

Q15 — Intersection of Two Lines

Q16 — Specific Term in a Binomial Expansion

Q17 — Vector Geometry Proof

Q18 — Closest Point on a Line to the Origin

Q19 — Combinatorics Problem Solving

Q20 — Induction with Inequality