2021 H2 Maths Paper 2 Q7 — Worked Solution

Part (a)

2 marks

Since the letters repeat, we divide the number of permutations of $11$ distinct objects by the factorial of each repeated letter’s count.

\text{no. of arrangements}

= \frac{11!}{5!\, 2!\, 2!}

= 83160

To check your answer:

11! = 39916800

and

5! \times 2! \times 2! = 480

;

39916800 \div 480 = 83160

, a positive integer as expected. ✓

$\therefore$ the number of different arrangements is $83160$ .

Part (b)

3 marks

Approach

each condition glues a group of letters together, so we treat each group as a single unit and arrange the units. The A’s need care: “exactly 4 of the 5 A’s together” is not just “glue 4 A’s into a block” — we must also make sure that block does not end up next to the 5th A, since that would merge into a run of all 5. We count all arrangements with a block of 4 A’s, then subtract those where the block sits next to the lone A.

Since the 2 B’s must be together, treat them as one unit $\lbrack BB\rbrack$ . Since the 2 R’s must be together, treat them as one unit $\lbrack RR\rbrack$ . Since C is next to D, treat them as one unit $\lbrack CD\rbrack$ , which has $2$ internal orders ( $CD$ or $DC$ ). Since exactly 4 of the A’s are together, treat those 4 as one unit $\left\lbrack \text{AAAA} \right\rbrack$ , leaving the 5th A as a separate letter.

This gives $5$ units to arrange in a row: $\lbrack BB\rbrack$ , $\lbrack RR\rbrack$ , $\lbrack CD\rbrack$ , $\left\lbrack \text{AAAA} \right\rbrack$ , $A$ .

\text{no. of arrangements of the}\ 5\ \text{units}

= 5! \times 2

= 240

From these $240$ , we subtract the arrangements in which $\left\lbrack \text{AAAA} \right\rbrack$ and the lone $A$ are adjacent, since that would give a run of $5$ A’s rather than exactly $4$ . Gluing $\left\lbrack \text{AAAA} \right\rbrack$ and $A$ together (in either order) leaves $4$ units to arrange:

\text{no. with all}\ 5\ A'\text{s together}

= 4! \times 2\ \left( \text{order of the glued A-block} \right) \times 2\ \left( CD\ \text{order} \right)

= 96

\text{no. of valid arrangements} = 240 - 96 = 144

To check your answer: a direct computer enumeration of all

83160

arrangements of ABRACADABRA, filtered by “BB adjacent”, “RR adjacent”, “C next to D” and “the A’s form a run of length exactly 4”, also gives

144

. ✓

$\therefore$ the number of such arrangements is $144$ .

Part (c)

3 marks

Treating all $5$ A’s as one unit $\left\lbrack \text{AAAAA} \right\rbrack$ , we have $7$ units to arrange: $\left\lbrack \text{AAAAA} \right\rbrack$ , $B$ , $B$ , $R$ , $R$ , $C$ , $D$ , with the B’s and R’s each repeated.

\text{no. of arrangements with all}\ 5\ A'\text{s together}

= \frac{7!}{2!\, 2!}

= 1260

Since all $83160$ arrangements from (a) are equally likely,

P\left( \text{all 5 A’s together} \right) = \frac{1260}{83160}

= \frac{1}{66}

To check your answer:

83160 \div 1260 = 66

exactly, so the fraction reduces to the unit fraction

\frac{1}{66}

. ✓

$\therefore$ $P\left( \text{all 5 A’s together} \right) = \frac{1}{66}$ .

2021 H2 Maths Paper 2, Question 7

Part (a)

Part (b)

Part (c)

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