Higher June 2025 Paper 6 Q18
18 Sasha creates a four-digit passcode.
They use four digits from 0 to 9.
They do not use any digit more than once.
Work out how many of the possible four-digit passcodes are even. [4]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 2520 | 4 | M3 for (10 × 9 × 8 × 7) ÷ 2 oe or M2 for 10 × 9 × 8 × 7 oe may be implied by 5040 | e.g. 5040 ÷ 2 or 5 × 9 × 8 × 7 For M2 they must be working in whole numbers not fractions apart from probabilities as below 5040 in body via whole numbers subsequently spoilt: M2 for 5040/n, otherwise M1 |
| or M1 for \(\frac{a}{10} \times \frac{b}{9} \times \frac{c}{8} \times \frac{d}{7}\) may be implied by \(\frac{k}{5040}\) or \(\frac{10}{a} \times \frac{9}{b} \times \frac{8}{c} \times \frac{7}{d}\) may be implied by \(\frac{5040}{k}\) where \(a\), \(b\), \(c\), \(d\) and \(k \ne 1\) | Condone answers and working in probabilities to a max of M3: e.g. M3 for answer \(\frac{1}{2520}\) e.g. M2 for \(\frac{1}{10} \times \frac{1}{9} \times \frac{1}{8} \times \frac{1}{7}\) or \(\frac{1}{5040}\) e.g. SC1 for \(\frac{1}{9} \times \frac{1}{8} \times \frac{1}{7} \times \frac{1}{6}\) [× 2] | ||
| If 0 scored, instead award SC1 for (10 × 10 × 10 × 10) ÷ 2 oe or for 9 × 8 × 7 × 6 [÷ 2] oe or for 9 × 9 × 8 × 7 [÷ 2] oe or for 10 × 9 × 8 × 5 or for 10 × 9 × 8 × 4 | |||