Systematic listing and the product rule

Work systematically: fix the shirt, then go through the ties.

White: WP, WS, WC

Blue: BP, BS, BC

Grey: GP, GS, GC

That is $3 \times 3 = 9$ combinations.

(a) An even number must end in 2 or 4. Put the biggest digits first and keep an even digit for the end: $5412$.

(b) Take each first digit in turn and pair it with the other three:

1: 12, 14, 15; 2: 21, 24, 25; 4: 41, 42, 45; 5: 51, 52, 54 (12 numbers).

Work systematically, fixing the first throw then the second:

HHH, HHT, HTH, HTT

THH, THT, TTH, TTT

There are $2 \times 2 \times 2 = 8$ outcomes.

Primes between 20 and 30: 23 and 29 (21, 25, 27 are odd but not prime).

Pair each letter with each prime:

L23, L29, U23, U29

Work systematically. Fix the first throw as H, then list the four ways for the other two:

HHH, HHT, HTH, HTT

Then repeat with T first:

THH, THT, TTH, TTT

There are $2 \times 2 \times 2 = 8$ outcomes.

Fix who comes first, then swap the other two.

Mohsin first: MYL, MLY

Yusuf first: YML, YLM

Luke first: LMY, LYM

That gives $3 \times 2 = 6$ outcomes.

Each choice can go with every other choice, so multiply:

$16 \times 5 \times 3 = 240$

Starter and main combinations: $5 \times 8 = 40$.

$240 \div 40 = 6$, which is a whole number.

So Jack could be correct, if there are 6 desserts.

Each of the 14 boys can be paired with any of the 12 girls.

$14 \times 12 = 168$ ways

Number of pairs of teams: $\frac{10 \times 9}{2} = 45$

Each pair plays 4 times: $45 \times 4 = 180$

Number of meals $= 5 \times 12 \times x = 60x$

$60x = 420$, so $x = 7$

Options for each digit (1 to 9):

- cube: 1, 8 → 2
- prime: 2, 3, 5, 7 → 4
- greater than 6: 7, 8, 9 → 3
- odd: 1, 3, 5, 7, 9 → 5

Number of passcodes $= 2 \times 4 \times 3 \times 5 = 120$

Probability $= \frac{1}{120}$

Guppy and angel: $7 \times 5 = 35$

Tetra and angel: $13 \times 5 = 65$

Guppy, tetra and angel: $7 \times 13 \times 5 = 455$

Total: $35 + 65 + 455 = 555$

Use the product rule: multiply the number of choices for each role.

$16 \times 120 \times 100 = 192\,000$

(a) Each dial has 5 choices: $5 \times 5 \times 5 = 125$

(b) The first dial has 5 choices, the second 4 left, the third 3 left: $5 \times 4 \times 3 = 60$

Total choices $= 19 \times r \times 25$

$19 \times 25 = 475$

$r = 3325 \div 475 = 7$ roof colours.

Starter and main: $9 \times 15 = 135$

Main and dessert: $15 \times 8 = 120$

All three: $9 \times 15 \times 8 = 1080$

Total: $135 + 120 + 1080 = 1335$

Each of the 16 teams plays the other 15 teams twice (once as each of the two sides), so every ordered pair of teams is one match.

Number of ordered pairs: $16 \times 15 = 240$

(Check: $\frac{16 \times 15}{2} = 120$ pairs, each playing twice, gives 240.)

The first student can be chosen in 30 ways and the second in 29 ways: $30 \times 29 = 870$.

Each pair has been counted twice (AB and BA), so $870 \div 2 = 435$.