The eight digits 1, 2, 2, 3, 4, 4, 4, 5 are arranged in a line.
- (a)How many different arrangements are there of these 8 digits?[1]
- (b)Find the number of different arrangements of the 8 digits in which there is a 2 at the beginning, a 2 at the end and the three 4s are not all together.[4]
- (c)Three digits are selected at random from the eight digits 1, 2, 2, 3, 4, 4, 4, 5. Find the probability that the three digits are all different.[5]
Mark scheme
Marking points from the official mark scheme (May/June 2024). The mark or mark type is shown in brackets.
- [B1] 8! / (2!3!) = 3360
- [M1] 6! / 3! – x r – s, r = 1, 2 and s a positive integer (including 0).
- [B1] 4! Seen either alone or in t – 4!, t an integer value > 24.
- [M1] 6! / 3! – x r – 4! x u, r =1, 2 and u = 1, 2.
- [A1] 96
- [M1] One correct calculation, unsimplified for an identified scenario containing 1, 2 and/or 1, 4.
- [A1] Two correct outcomes evaluated, accept unsimplified.
- [M1] Four correct scenarios added.
- [B1] Used as denominator of probability expression. C(8,3) [= 56]
- [A1] 34/56 = 17/28 = 0.607
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