Date | May 2017 | Marks available | 6 | Reference code | 17M.2.hl.TZ2.8 |
Level | HL only | Paper | 2 | Time zone | TZ2 |
Command term | Find | Question number | 8 | Adapted from | N/A |
Question
In a trial examination session a candidate at a school has to take 18 examination papers including the physics paper, the chemistry paper and the biology paper. No two of these three papers may be taken consecutively. There is no restriction on the order in which the other examination papers may be taken.
Find the number of different orders in which these 18 examination papers may be taken.
Markscheme
METHOD 1
consideration of all papers
all papers may be sat in 18! ways A1
number of ways of positioning “pairs” of science subjects \( = 3! \times 17!\) A1
but this includes two copies of each “triple” (R1)
number of ways of positioning “triplets” of science subjects \( = 3! \times 16!\) A1
hence number of arrangements is \(18! - 3! \times 17! + 3! \times 16!\) M1A1
\(( = 4.39 \times {10^{15}})\)
METHOD 2
consideration of all the non-science papers (M1)
hence all non-science papers can be sat in 15! ways A1
there are \(16 \times 15 \times 14{\text{ }}( = 3360)\) ways of positioning the three science papers (M1)A1
hence the number of arrangements is \(16 \times 15 \times 14 \times 15!{\text{ (}} = 4.39 \times {10^{15}})\) (M1)A1
METHOD 3
consideration of all papers
all papers may be sat in 18! ways A1
number of ways of positioning exactly two science subjects \( = 3! \times 15! \times 16 \times 15\) M1A1
number of ways of positioning “triplets” of science subjects \( = 3! \times 16!\) A1
hence number of arrangements is \(18! - 3! \times 16! - 3! \times 15! \times 16 \times 15\) M1A1
\(( = 4.39 \times {10^{15}})\)
[6 marks]