2016 H2 Maths Paper 2 Q7 — Worked Solution

The board has $6$ men and $4$ women, $10$ members in total. A chairperson, a secretary and a treasurer are three different roles, so choosing who fills them is an arrangement (order matters), not just a selection.

Part (i)

1 mark

Since all three roles are filled by women, we are arranging $3$ of the $4$ women into $3$ distinct roles: $4$ choices for chairperson, then $3$ for secretary, then $2$ for treasurer.

\text{no. of ways} = 4 \times 3 \times 2 = 24

$\therefore$ there are $24$ ways.

Part (ii)

3 marks

Approach

“At least one is a woman and at least one is a man” is easiest by the complement: from all ways of filling the $3$ roles, remove the ways that are all men and the ways that are all women (these two cases are mutually exclusive and together are the complement of “at least one of each”).

The total number of ways to fill the $3$ roles from all $10$ members is

10 \times 9 \times 8 = 720

The number of ways with all $3$ roles filled by men is

6 \times 5 \times 4 = 120

and from part (i), the number of ways with all $3$ roles filled by women is $24$ .

\text{no. of ways} = 720 - 120 - 24 = 576

$\therefore$ there are $576$ ways.

Part (iii)

3 marks

Approach

The $10$ members sit at random around a (circular) table, so we count circular arrangements, where only rotations of the same seating are identified (reflections are distinct, since swapping left and right neighbours gives a genuinely different seating). The total number of such arrangements of $10$ distinct people is $(10 - 1)! = 9!$ . For the favourable outcomes, we treat the chairperson, secretary and treasurer as a single block, since they must sit in three adjacent places.

Treating the chairperson, secretary and treasurer as one block leaves $8$ objects (the block and the other $7$ members) to arrange around the table:

\text{no. of arrangements of the }8\text{ objects} = (8 - 1)! = 7!

Within the block, the $3$ officers can be ordered in $3!$ ways.

P\left( \text{3 officers adjacent} \right) = \frac{7! \times 3!}{9!}

= \frac{7! \times 6}{9 \times 8 \times 7!}

= \frac{6}{72}

= \frac{1}{12}

$\therefore\ P\left( \text{chairperson, secretary and treasurer sit in three adjacent places} \right) = \frac{1}{12}$ .

Part (iv)

3 marks

Approach

“Separated from each other by at least one other person” means no two of the three officers are adjacent. Rather than working with the officers directly, we first seat the other $7$ members around the table, which creates $7$ gaps between consecutive members (a circular arrangement of $7$ people has $7$ such gaps); placing each officer into a different gap then guarantees no two officers are next to each other.

Seat the other $7$ members around the table first:

\text{no. of arrangements of the }7\text{ members} = (7 - 1)! = 6!

This creates $7$ gaps (one between each pair of adjacent members). Choose $3$ of these $7$ gaps for the $3$ officers, one officer per gap, and arrange the officers:

\text{no. of ways to place the officers} = \binom{7}{3} \times 3!

P\left( \text{no two officers adjacent} \right) = \frac{6! \times \binom{7}{3} \times 3!}{9!}

= \frac{720 \times 35 \times 6}{362880}

= \frac{5}{12}

To check your answer:

P\left( \text{adjacent} \right) + P\left( \text{exactly one other person between some pair, etc.} \right) + P\left( \text{all separated} \right) = 1

; here

\frac{1}{12} + \frac{5}{12} = \frac{1}{2}

, leaving

\frac{1}{2}

for the remaining case (exactly one pair adjacent, the third officer elsewhere), which is a sensible split. ✓

$\therefore\ P\left( \text{chairperson, secretary and treasurer all separated by at least one other person} \right) = \frac{5}{12}$ .

2016 H2 Maths Paper 2, Question 7

Part (i)

Part (ii)

Part (iii)

Part (iv)

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