RI 2025 H2 Math Prelim Paper 2 — Permutations & Combinations (worked solution)

\textbf{(a)} $2880$ ways \textbf{(b)} $86400$ ways \textbf{(c)} Total $= 3600 + 151200 + 14400 = \mathbf{169200}$ ways

Permutations & Combinations: Charms: circular; keychain — RI 2025 H2 Math Prelim Paper 2

Raffles Institution2025 PrelimPaper 2●●● Challenging11 marks

What this question tests

Charms: circular; keychain.

Question

Kitty has 5 small, 3 medium and 2 large spherical charms and each of the 10 charms is uniquely designed.

Kitty arranges all the charms in a circle on a corkboard with the charms of the same sizes next to each other. How many ways are there for her to do so?

On another occasion, Kitty arranges all the charms in a line at the base of a photo frame with none of the small charms next to each other. How many ways are there for her to do so?

The diameters of a small, medium and large charm are 0.5 cm, 1 cm and 2 cm respectively. Each of the spherical charms has a ring that allows it to be threaded on the chain.

Kitty makes a keychain that includes a 6 cm chain with one end attached to a keyring, as shown in Fig. 1. It is given that the 6 cm chain is fully threaded with the charms, with no space between them, and is strung together in a similar manner to Fig. 2. Fig. 2 shows the 6 cm chain threaded with 2 large and 2 medium charms.

Figure for this question

How many different ways can she make a keychain with at least one charm of each of the three sizes?

How many different ways can she choose 3 objects from the 10 charms to arrange in a row? In your answer, state the cases you are considering.

Show full worked solution▾

Arrange 3 groups (5 small, 3 medium, 2 large) in a circle. First arrange the 3 groups in a circle: $(3-1)!=2$ ways. Then arrange within groups: $5!\cdot 3!\cdot 2!$ ways. \[ \text{Total} = 2!\times 5!\times 3!\times 2! = 2\times 120\times 6\times 2 = \mathbf{2880} \]

Place 5 non-small charms in a line: $5!$ ways. Insert 5 small charms into the 6 gaps: $\binom{6}{5}\times 5!$ ways. \[ \text{Total} = 5!\times\binom{6}{5}\times 5! = 120\times 6\times 120 = \mathbf{86400} \]

Chain is 6 cm. Diameters: small $=0.5$ cm, medium $=1$ cm, large $=2$ cm. Need at least one of each size, total length exactly 6 cm: \[ 0.5s + m + 2l = 6, \quad s\geq 2\text{ (must be even)},\; m\geq 1,\; l\geq 1 \]

Valid combinations $(l,m,s)$:

$l=2, m=1, s=2$: $4+1+1=6$ cm. $s=2\leq 5$, $m=1\leq 3$, $l=2\leq 2$.
$l=1, m=2, s=4$: $2+2+2=6$ cm.
$l=1, m=3, s=2$: $2+3+1=6$ cm.

Chain is linear with one end fixed at keyring. Choose and arrange:

Case 1 ($2L, 1M, 2S$): $\binom{2}{2}\binom{3}{1}\binom{5}{2}\times 5! = 1\times 3\times 10\times 120 = 3600$

Case 2 ($1L, 2M, 4S$): $\binom{2}{1}\binom{3}{2}\binom{5}{4}\times 7! = 2\times 3\times 5\times 5040 = 151200$

Case 3 ($1L, 3M, 2S$): $\binom{2}{1}\binom{3}{3}\binom{5}{2}\times 6! = 2\times 1\times 10\times 720 = 14400$

Answer: (a) $2880$ ways (b) $86400$ ways (c) Total $= 3600 + 151200 + 14400 = \mathbf{169200}$ ways

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