Practice Maths

Solutions — Combinations

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  1. Q1 — Calculate Combinations

    (a) C(7, 2) = (7 × 6) / (2 × 1) = 42 / 2 = 21

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

    (c) C(9, 9) = 9! / (9! × 0!) = 1 / 1 = 1 (there is only one way to choose all items)

    (d) C(12, 10) = C(12, 2) by symmetry (since 12 − 10 = 2) = (12 × 11) / 2 = 66

  2. Q2 — Choosing a Committee

    No roles are assigned, so order does not matter. Apply the combination formula with n = 9, r = 4:

    C(9, 4) = 9! / (4! × 5!) = (9 × 8 × 7 × 6) / (4 × 3 × 2 × 1) = 3024 / 24 = 126 committees

  3. Q3 — Selecting Players

    We select 11 starters from 15; no specific positions are assigned, so order does not matter → C(15, 11).

    Use symmetry: C(15, 11) = C(15, 4) = (15 × 14 × 13 × 12) / (4 × 3 × 2 × 1) = 32760 / 24 = 1 365 ways

    Using the symmetry property C(n,r) = C(n,n−r) significantly reduces computation here.

  4. Q4 — Combinations vs Permutations

    (a) Debate team — three representatives with no distinct roles, so order irrelevant → C(10, 3).

    C(10, 3) = (10 × 9 × 8) / (3 × 2 × 1) = 720 / 6 = 120 ways

    (b) Three distinct leadership roles — different roles make order matter → P(10, 3).

    P(10, 3) = 10 × 9 × 8 = 720 ways

    The ratio 720 / 120 = 6 = 3!, confirming that permutations count each combination 3! = 6 times (once for each ordering of the same 3 people).

  5. Q5 — Mixed Gender Selection

    (a) Exactly 3 boys AND 2 girls:

    Choose 3 from 6 boys: C(6, 3) = (6 × 5 × 4)/(3 × 2 × 1) = 120/6 = 20.

    Choose 2 from 5 girls: C(5, 2) = (5 × 4)/(2 × 1) = 10.

    Total = 20 × 10 = 200 teams

    (b) Exactly 4 girls AND 1 boy:

    Choose 4 from 5 girls: C(5, 4) = C(5, 1) = 5.

    Choose 1 from 6 boys: C(6, 1) = 6.

    Total = 5 × 6 = 30 teams

  6. Q6 — Card Hands

    (a) Exactly 3 aces:

    Choose 3 aces from 4: C(4, 3) = 4.

    Choose the remaining 2 cards from the 48 non-aces: C(48, 2) = (48 × 47)/2 = 1128.

    Total = 4 × 1128 = 4 512 hands

    (b) All 5 cards from one suit (flush):

    Choose the suit: C(4, 1) = 4.

    Choose 5 cards from 13 in that suit: C(13, 5) = (13 × 12 × 11 × 10 × 9)/(5!) = 1287.

    Total = 4 × 1287 = 5 148 flushes

  7. Q7 — Committees with a Required Member

    Person A is fixed on the committee. This reduces the problem to choosing 4 more members from the remaining 11 people.

    C(11, 4) = (11 × 10 × 9 × 8) / (4 × 3 × 2 × 1) = 7920 / 24 = 330 committees

  8. Q8 — Committees with an Excluded Pair

    Complementary counting: Valid = Total − Both on committee.

    Total unrestricted committees: C(10, 4) = (10 × 9 × 8 × 7)/(4 × 3 × 2 × 1) = 210.

    Committees with both Tom AND Jerry: fix both; choose 2 more from remaining 8.

    Invalid = C(8, 2) = (8 × 7)/2 = 28.

    Valid = 210 − 28 = 182 committees

  9. Q9 — Selecting at Least One

    Complementary counting: At least one teacher = Total − No teachers.

    Total committees from 10 people: C(10, 4) = 210.

    Committees with no teachers (all 4 from 6 students): C(6, 4) = C(6, 2) = 15.

    At least one teacher = 210 − 15 = 195 committees

    Verification by direct count:

    1 teacher + 3 students: C(4,1) × C(6,3) = 4 × 20 = 80

    2 teachers + 2 students: C(4,2) × C(6,2) = 6 × 15 = 90

    3 teachers + 1 student: C(4,3) × C(6,1) = 4 × 6 = 24

    4 teachers + 0 students: C(4,4) = 1

    Total = 80 + 90 + 24 + 1 = 195. ✓

  10. Q10 — Handshakes in a Group

    Each handshake involves a unique unordered pair of people, so the number of handshakes is C(n, 2).

    C(n, 2) = n(n−1)/2 = 45

    Multiply both sides by 2: n(n−1) = 90

    This is a quadratic: n2 − n − 90 = 0 → (n−10)(n+9) = 0

    Since n must be positive: n = 10.

    There are 10 people at the meeting.