O-LevelAdditional MathematicsPermutations and combinations4037/21 · May/June 2022 · Q614 Marks

A 5-digit number is to be formed from the seven digits 0, 1, 2, 3, 4, 5, 6. Each digit can be used at most once in any number and the number does not start with 0. Find the number of ways in which this can be done.

(a)(i)A 5-digit number is to be formed from the seven digits 0, 1, 2, 3, 4, 5, 6. Each digit can be used at most once in any number and the number does not start with 0. Find the number of ways in which this can be done.[2]
(a)(ii)Find how many of these 5-digit numbers are even.[3]
(b)A team of 7 people is to be selected from a group of 9 women and 6 men. Find the number of different teams that can be selected which include at least one man.[2]
(c)(i)Show that ⁿC₃+ⁿC₂ = (1/6)(n³-n) for n ≥ 3.[5]
(c)(ii)Hence solve the equation ⁿC₃ + ⁿC₂ = 4n where n ≥ 3.[2]

Mark scheme

Marking points from the official mark scheme (May/June 2022). The mark or mark type is shown in brackets.

[M1] 6×6×5×4 × 3 oe
[A1] 2160
[M2] Full correct calculation (360 + 900) ('how many end with 0' + 'how many end with 2, 4, 6' oe) (M1 for any correct product soi Eg (6 × 5 × 4 × 3 × 1) or 360 (5×5×4×3× 1) or 300 (5×5×4×3× 3) or 900)
[A1] 1260 cao
[M1] 15C7 - 9C7
[A1] 6399
[B2] n!/(n-3)!3! + n!/(n-2)!2! (B1 for either expression correct)
[M2] n(n-1)(n-2)/6 + n(n-1)/2 (M1 for either expression correct)
[A1] n(n-1)(n-2+3)/6 or n³-3n²+2n/6 + 3n²-3n/6 leading to 1/6(n³-n)
[M1] n(n² – 25) = 0 oe
[A1] n = 5 as the only solution

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