2015 H2 Maths Paper 2 Q11 — Worked Solution

The word CABBAGES has $8$ letters: $A$ (twice), $B$ (twice), and $C$ , $G$ , $E$ , $S$ (once each).

Part (i)

2 marks

\text{no. of arrangements} = \frac{8!}{2!\, 2!}

= 10080

$\therefore$ the number of different arrangements is $10080$ .

Part (ii)

1 mark

10080

Explanation: among all $10080$ arrangements, exactly one has its letters in alphabetical order — namely AABBCEGS itself. Every other arrangement has the letters out of order, so the number of arrangements not in alphabetical order is $10080 - 1 = 10079$ .

$\therefore$ the number of arrangements not in alphabetical order is $10079$ .

Part (iii)

2 marks

Approach

treat the two A’s as a single block $\lbrack AA\rbrack$ and the two B’s as a single block $\lbrack BB\rbrack$ . This leaves $6$ distinguishable entities to arrange in a row: $\lbrack AA\rbrack$ , $\lbrack BB\rbrack$ , $C$ , $G$ , $E$ , $S$ , with no repeats among them.

\text{no. of arrangements} = 6!

= 720

$\therefore$ the number of arrangements with both A’s together and both B’s together is $720$ .

Part (iv)

4 marks

Approach

we want to exclude arrangements where the two A’s are adjacent, and arrangements where the two B’s are adjacent (C, G, E, S each occur only once, so they can never repeat adjacently). This is a “not both” count, done by inclusion-exclusion on the two events “A’s together” and “B’s together”.

If the A’s are together, treat them as one block $\lbrack AA\rbrack$ : this leaves $7$ entities — $\lbrack AA\rbrack$ , $B$ , $B$ , $C$ , $G$ , $E$ , $S$ — with the two B’s still repeated.

\text{no. with A}'\text{s together} = \frac{7!}{2!}

= 2520

By the same argument with the roles of $A$ and $B$ swapped,

\text{no. with B}'\text{s together} = \frac{7!}{2!}

= 2520

From part (iii), the number of arrangements with both pairs together is $720$ .

By inclusion-exclusion,

\text{no. with A}'\text{s together or B}'\text{s together} = 2520 + 2520 - 720

= 4320

\text{no. with no two adjacent letters the same} = 10080 - 4320

= 5760

To check your answer: a direct computer enumeration of all

10080

distinct arrangements of CABBAGES, counting those with no two equal letters adjacent, also gives

5760

. ✓

$\therefore$ the number of arrangements with no two adjacent letters the same is $5760$ .

2015 H2 Maths Paper 2, Question 11

Part (i)

Part (ii)

Part (iii)

Part (iv)

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