Combinatorics — Topic Review — Solutions

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This review covers all five lessons in the Combinatorics topic: Multiplication and Addition Principles, Permutations, Combinations, Pascal’s Triangle and the Binomial Theorem, and Inclusion–Exclusion and Pigeonhole. Questions are exam-style and increase in difficulty. Click each question to reveal the full worked solution.

Mixed Review Questions

1. Fluency

   Q1 — Meal Combinations (Multiplication Principle)

   A set menu offers 3 entrees, 5 mains, and 4 desserts. How many different 3-course meals can be ordered?

   ▶ Show Answer

   Each course is an independent sequential choice → multiply (multiplication principle).

   Total = 3 × 5 × 4 = 60 meals

2. Fluency

   Q2 — Factorial Simplification

   Simplify without a calculator:   (a) 9!/7!    (b) 12!/(3! × 9!)

   ▶ Show Answer

   (a) 9!/7! = (9 × 8 × 7!) / 7! = 9 × 8 = 72

   (b) 12!/(3! × 9!) = (12 × 11 × 10 × 9!) / (6 × 9!) = (12 × 11 × 10) / 6 = 1320 / 6 = 220

   Note: 12!/(3!×9!) = C(12, 3) = 220.

3. Fluency

   Q3 — Arranging Books on a Shelf (Permutations)

   In how many ways can 5 different books be arranged on a shelf?

   ▶ Show Answer

   Arranging all 5 distinct books in 5 positions: order matters → P(5, 5) = 5!

   5! = 5 × 4 × 3 × 2 × 1 = 120 arrangements

4. Fluency

   Q4 — Choosing a Team (Combinations)

   In how many ways can a group of 3 be chosen from 8 people to serve on a committee (no assigned roles)?

   ▶ Show Answer

   No roles assigned, order does not matter → C(8, 3).

   C(8, 3) = (8 × 7 × 6) / (3 × 2 × 1) = 336 / 6 = 56 committees

5. Fluency

   Q5 — Binomial Expansion (Pascal’s Triangle)

   Expand (x + 3)³ using the binomial theorem.

   ▶ Show Answer

   Row 3 coefficients from Pascal’s triangle: 1, 3, 3, 1.

   (x + 3)³ = C(3,0)x³ + C(3,1)x²(3) + C(3,2)x(3)² + C(3,3)(3)³

   = x³ + 3 × 3 × x² + 3 × 9 × x + 27

   = x³ + 9x² + 27x + 27

6. Fluency

   Q6 — Survey Data (Inclusion–Exclusion)

   Of 60 people surveyed, 35 drink coffee, 28 drink tea, and 14 drink both. How many drink at least one of the two beverages?

   ▶ Show Answer

   |C ∪ T| = |C| + |T| − |C ∩ T| = 35 + 28 − 14 = 49 people

   People who drink neither = 60 − 49 = 11.

7. Understanding

   Q7 — Seating with a Restriction (Permutations)

   Seven people sit in a row. In how many ways can this be done if two particular people, Alex and Sam, must NOT sit next to each other?

   ▶ Show Answer

   Complementary counting: Valid = Total − (Alex and Sam adjacent).

   Total arrangements of 7 people: 7! = 5040.

   Arrangements where Alex and Sam ARE adjacent: treat them as one unit → 6 units.

   Arrangements of 6 units: 6! = 720. Alex and Sam can swap within the pair: × 2.

   Adjacent arrangements = 6! × 2 = 1440.

   Valid = 5040 − 1440 = 3 600 arrangements

8. Understanding

   Q8 — Selecting a Mixed Group (Combinations)

   A committee of 6 is chosen from 8 men and 5 women. How many committees contain exactly 4 men and 2 women?

   ▶ Show Answer

   Choose 4 men from 8: C(8, 4) = (8 × 7 × 6 × 5)/(4 × 3 × 2 × 1) = 1680/24 = 70.

   Choose 2 women from 5: C(5, 2) = (5 × 4)/2 = 10.

   Total = 70 × 10 = 700 committees

9. Understanding

   Q9 — Specific Term in a Binomial Expansion

   Find the coefficient of x² in the expansion of (3x − 2)⁵.

   ▶ Show Answer

   Write as (3x + (−2))⁵. General term: T_r+1 = C(5, r)(3x)^5−r(−2)^r = C(5, r)  3^5−r  (−2)^r  x^5−r

   For x²: set 5 − r = 2, so r = 3.

   T₄ = C(5, 3) × 3² × (−2)³ × x²

   = 10 × 9 × (−8) × x²

   = −720 × x²

   The coefficient of x² is −720.

10. Understanding

    Q10 — Three Subjects (Inclusion–Exclusion)

    In a group of 200 students: 110 study Biology, 85 study Chemistry, 70 study Physics. Pairwise: 40 study Bio and Chem, 30 study Bio and Phys, 25 study Chem and Phys. 15 study all three. How many students study none of the three subjects?

    ▶ Show Answer

    |B ∪ C ∪ P| = |B| + |C| + |P| − |B∩C| − |B∩P| − |C∩P| + |B∩C∩P|

    = 110 + 85 + 70 − 40 − 30 − 25 + 15

    = 265 − 95 + 15

    = 185

    Students who study none of the three = 200 − 185 = 15 students

11. Understanding

    Q11 — Arrangements with Repeated Letters (Permutations)

    How many distinct arrangements are there of the letters in STATISTICS?

    ▶ Show Answer

    STATISTICS has 10 letters: S(3), T(3), A(1), I(2), C(1).

    Arrangements = 10! / (3! × 3! × 2!) = 3 628 800 / (6 × 6 × 2) = 3 628 800 / 72 = 50 400

12. Problem Solving

    Q12 — At Least Restriction (Combinations)

    A hand of 5 cards is dealt from a standard 52-card deck. How many hands contain at least one ace? (There are 4 aces in the deck.)

    ▶ Show Answer

    Complementary counting: At least one ace = Total − No aces.

    Total 5-card hands: C(52, 5) = (52 × 51 × 50 × 49 × 48) / 120 = 2 598 960.

    Hands with no aces (all 5 from 48 non-aces): C(48, 5) = (48 × 47 × 46 × 45 × 44) / 120.

    Numerator: 48 × 47 = 2256; × 46 = 103776; × 45 = 4 669 920; × 44 = 205 476 480.

    C(48, 5) = 205 476 480 / 120 = 1 712 304.

    At least one ace = 2 598 960 − 1 712 304 = 886 656 hands

13. Problem Solving

    Q13 — Constant Term in a Binomial Expansion

    Find the constant term in the expansion of (2x² + 1/x)⁹.

    ▶ Show Answer

    Write as (2x² + x⁻¹)⁹. General term:

    T_r+1 = C(9, r)(2x²)^9−r(x⁻¹)^r = C(9, r)  2^9−r  x^2(9−r)  x^−r = C(9, r)  2^9−r  x^18−2r−r = C(9, r)  2^9−r  x^18−3r

    For a constant term: 18 − 3r = 0 ⇒ r = 6.

    T₇ = C(9, 6) × 2³ × x⁰ = C(9, 3) × 8 = 84 × 8 = 672

14. Problem Solving

    Q14 — Pigeonhole Application

    A sequence of 101 integers is chosen from {1, 2, 3, …, 200}. Prove that at least two of the integers in the sequence sum to 201.

    ▶ Show Answer

    Construction of containers: Partition {1, 2, …, 200} into 100 pairs, each summing to 201:

    {1, 200}, {2, 199}, {3, 198}, …, {100, 101}.

    There are exactly 100 such pairs (containers). We choose 101 integers (objects) from these 200 values.

    By the pigeonhole principle, since 101 > 100, at least two of the chosen integers must come from the same pair.

    Two integers from the same pair sum to 201.

    Therefore, at least two integers in the sequence sum to 201. □

15. Problem Solving

    Q15 — Combined Counting Problem

    A password is 6 characters long. It must contain at least one digit (0–9) and at least one letter (A–Z, 26 letters). Characters may repeat, and the order matters. How many valid passwords are there? (Hint: use complementary counting.)

    ▶ Show Answer

    The character set has 26 letters + 10 digits = 36 characters.

    Total passwords (unrestricted, any 6 characters from 36): 36⁶.

    Complementary counting: Valid = Total − (No digits) − (No letters).

    Passwords with NO digits (all 6 characters are letters): 26⁶.

    Passwords with NO letters (all 6 characters are digits): 10⁶.

    These two sets are mutually exclusive (a password cannot simultaneously have no digits and no letters unless it is empty, which is impossible here).

    Valid passwords = 36⁶ − 26⁶ − 10⁶

    36⁶ = 2 176 782 336

    26⁶ = 308 915 776

    10⁶ = 1 000 000

    Valid = 2 176 782 336 − 308 915 776 − 1 000 000 = 1 866 866 560 passwords