2017 H2 Maths Paper 2 Q6 — Worked Solution

Part (i)

2 marks

Approach

Treat each family’s 4 cards as a single block, since the 4 cards in each family set must stay next to each other. There are 5 such blocks to arrange, and within each block the 4 family members can be ordered independently.

\text{no. of ways} = 5! \times (4!)^{5}

= 120 \times 24^{5}

= 955\, 514\, 880

$\therefore$ the cards can be arranged in $955\, 514\, 880$ ways.

Part (ii)

3 marks

Approach

Father Red belongs to both the father block (5 fathers) and the Red family block (4 Red cards), and Father Blue belongs to both the father block and the Blue family block. Since Father Red is the only card common to the father block and the Red block, these two blocks can each be a run of consecutive cards only if they overlap at that one shared card — i.e. Father Red sits at one end of the father block, with the other 3 Red family members running on immediately outside that end. The same argument applies to Father Blue at the other end of the father block. So all three conditions force a single mega-block of $5 + 3 + 3 = 11$ consecutive cards (5 fathers, plus the 3 non-father Red cards, plus the 3 non-father Blue cards), with the remaining 9 cards (the non-father members of the Green, Yellow and Orange families) free to sit anywhere else in the row.

Within the mega-block:

Father Red and Father Blue must occupy the two ends of the 5-card father run (they cannot both be at the same end), so there are $2$ ways to decide which end is Red and which is Blue.
The remaining 3 fathers (Green, Yellow, Orange) fill the middle of the father run, in $3!$ ways.
The 3 non-father Red cards run outward from Father Red, in $3!$ ways.
The 3 non-father Blue cards run outward from Father Blue, in $3!$ ways.

\text{arrangements within the mega-block} = 2 \times 3! \times 3! \times 3! = 432

Treating the mega-block as one unit, we now arrange it together with the other $9$ free cards: $10$ units in total.

\text{no. of ways} = 10! \times 432

= 3\, 628\, 800 \times 432

= 1\, 567\, 641\, 600

To check your answer: on a scaled-down version of this problem (3 families of 2 members: father and mother, fathers together, and two of the three family-pairs each together), the same overlapping-block argument gives

2! \times 2 = 4

, which matches a direct brute-force count of all

6! = 720

arrangements. ✓

$\therefore$ the cards can be arranged in $1\, 567\, 641\, 600$ ways.

Part (iii)

4 marks

Approach

Arranging the 15 non-father cards around the circle first leaves 15 gaps between them (a circular arrangement of $n$ objects has $n$ gaps). Placing the 5 distinct father cards into 5 different gaps then guarantees no two father cards are adjacent, since each gap holds at most one father.

There are $20$ distinct cards, so the total number of circular arrangements is

(20 - 1)! = 19!

Arrange the $15$ non-father cards in a circle first:

(15 - 1)! = 14!\ \text{ways}

This creates $15$ gaps between consecutive non-father cards. Choosing and arranging the $5$ distinct father cards into $5$ of these $15$ gaps (order matters, since the fathers are distinct cards):

\frac{15!}{10!}\ \text{ways}

P\left( \text{no two father cards adjacent} \right) = \frac{14! \times \frac{15!}{10!}}{19!}

= \frac{14 \times 13 \times 12 \times 11}{19 \times 18 \times 17 \times 16}

= \frac{1001}{3876}

= 0.258\ \left( 3\ \text{s.f.} \right)

To check your answer: by GC,

1001/3876 = 0.25826\ldots

, consistent with the exact fraction. ✓

$\therefore$ the probability that no two father cards are next to each other is $\frac{1001}{3876} = 0.258$ (3 s.f.).

2017 H2 Maths Paper 2, Question 6

Part (i)

Part (ii)

Part (iii)

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