Product counting and estimating powers/roots

(a) $676\,000$

1. Multiply the number of choices for each position.
   $$26 \times 26 \times 10 \times 10 \times 10 = 676\,000$$

• M1 Multiplying the choices, such as $26^2 \times 10^3$.
• A1 The correct answer, $676\,000$.

(b) $650\,000$

1. The second letter has only 25 choices.
2. $$26 \times 25 \times 1000 = 650\,000$$

• M1 Using $26 \times 25$ for the letters.
• A1 The correct answer, $650\,000$.

(a) $72$

1. $4 \times 6 \times 3 = 72$.

• B1 The correct answer, $72$.

(b) $42$

1. Main and starter: $6 \times 4 = 24$. Main and dessert: $6 \times 3 = 18$.
2. The two cases cannot both happen, so add: $24 + 18 = 42$.

• M1 Finding one case: 24 or 18.
• A1 The correct answer, $42$.

(a) $140$

1. $$5\times 4\times 7$$
2. The first letter has 5 choices and the second has 4 because repetition is not allowed.
3. The digit has 7 choices. The product rule gives 5 $\times$ 4 $\times$ 7 = 140 labels.

• P1 Establishing $5\times 4\times 7$ or an equivalent valid method.
• A1 Correct answer: $140$

(a) 8.6

1. $$8.6^{2}$$
2. $$8.7^{2}$$
3. For positive numbers, squaring preserves their order.
4. $8.6^{2}$ = 73.96 < 75, but $8.7^{2}$ = 75.69 > 75.
5. Therefore 8.6 < $\sqrt{75}$ < 8.7, so the required number is 8.6.

• P1 Establishing $8.6^{2}$ or an equivalent valid method.
• P1 Establishing $8.7^{2}$ or an equivalent valid method.
• C1 Correct conclusion with the complete supporting argument: 8.6

(a) $328$

1. Split by the last digit, because 0 affects the first digit.
   $$\text{last digit } 0: \ 9 \times 8 \times 1 = 72$$
2. Last digit 2, 4, 6 or 8: the first digit can be any of 1 to 9 except the last digit (8 choices), and the middle any of the 8 digits left.
   $$4 \times 8 \times 8 = 256$$
3. $$72 + 256 = 328$$

• P1 Splitting into cases: last digit 0, and last digit 2, 4, 6 or 8.
• P1 Counting the last-digit-0 case correctly: 72.
• P1 Counting the other case correctly: $4 \times 8 \times 8 = 256$.
• A1 The correct answer, $328$.

(a) $180$

1. $$7\times 6\times 5$$
2. $$3\times 2\times 5$$
3. Ignoring the restriction, the product rule gives 7 $\times$ 6 $\times$ 5 = 210 assignments.
4. If Isha and Leon both hold roles, choose the third person in 5 ways and arrange the three people in 3 $\times$ 2 $\times$ 1 = 6 ways: 30 forbidden assignments.
5. Subtract these assignments: 210 $-$ 30 = 180.

• P1 Establishing $7\times 6\times 5$ or an equivalent valid method.
• P1 Establishing $3\times 2\times 5$ or an equivalent valid method.
• A1 Correct answer: $180$

(a) $30$

1. If the last digit is zero there are four then three choices.
   $$4\times 3$$
2. For either non-zero even last digit there are three then three choices.
   $$2\times 3\times 3$$
3. Add the disjoint cases.
   $$12+18$$
4. Therefore $30$.

• P1 If the last digit is zero there are four then three choices.
• P1 For either non-zero even last digit there are three then three choices.
• P1 Add the disjoint cases.
• A1 Correct answer: $30$

(b) When zero is last it cannot also be first. When 2 or 4 is last, zero remains available but must be excluded from the first position.

1. When zero is last it cannot also be first. When 2 or 4 is last, zero remains available but must be excluded from the first position.

• C1 Correct conclusion with supporting reasoning: When zero is last it cannot also be first. When 2 or 4 is last, zero remains available but must be excluded from the first position.

(a) $144$

1. $$3\times 24$$
2. $$3\times 18$$
3. $$1\times 18$$
4. Divisibility by 4 depends on the last two digits. With these distinct digits, the possible endings are 04, 12, 20, 24, 32, 40 and 52.
5. For the 3 endings containing 0, the other four digits can be arranged in 4 $\times$ 3 $\times$ 2 $\times$ 1 = 24 ways each.
6. For the other 4 endings, 0 is among the remaining digits. There are 3 choices for the first digit and 3 $\times$ 2 $\times$ 1 choices for the others, giving 18 ways each.
7. Total = 3 $\times$ 24 + 4 $\times$ 18 = 144.

• P1 Establishing $3\times 24$ or an equivalent valid method.
• P1 Establishing $3\times 18$ or an equivalent valid method.
• P1 Establishing $1\times 18$ or an equivalent valid method.
• A1 Correct answer: $144$