OCR GCE Further Mathematics A Y534/01: Discrete Mathematics Advanced Subsidiary GCE Mark Scheme for Autumn 2021

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Mark Scheme October 2021 Question Answer Marks AO Guidance 1 (a) 5 partitions into a set of size 1 and a set of size 4 B1 1.1 5 where smaller set has size 1 or 5C1 = 5 {X | X, X, X, X} 10 partitio... ns into a set of size 2 and a set of size 3 B1 2.5 10 where smaller set has size 2, with an explanation of why it is 10 {X, X | X, X, X} because there are 5C2 choices for the set of size 2 (note the totalof 15 is given in the question) e.g. 5C2 = 10 or (5×4) ÷ 2 = 10 or 4 + 3 + 2 + 1 = 10 Alternative solution {A}, {B, C, D, E} {B}, {A, C, D, E} B1 List (or any equivalent) that has exactly 5 distinct cases where {C}, {A, B, D, E} {E}, {A, B, C, D} {D}, {A, B, C, E} smaller set has size 1 May just list one set, e.g. {A}, {B}, {C}, {D}, {E} {A, B}, {C, D, E} {A, C}, {B, D, E} List (or any equivalent) that has 10 distinct cases sets where smaller {A, D}, {B. C. E} {B, C}, {A, D, E} {B, E}, {A, C, D} {C, E}, {A, B, D} {A, E}, {B, C, D} {B, D}, {A, C, E} {C, D}, {A, B, E} {D, E}, {A, B, C} B1 set has size 2 May just list one set, e.g. {A, B}, {A, C} {A, D}, {A, E}, {B, C}, {B, D}, {B, E}, {C, D}, {C, E}, {D, E} [2] 1 (b) Partitions into sets of sizes 1, 1 and 3 M1 1.1 Considering cases where set sizes are 1, 1, 3 5 × 4 ÷ 2 = 10 partitions of this type A1 2.1 Explanationof why there are 10 of these e.g. 5C3 = 10 or 5 × 4 ÷ 2 = 10 or a list of the cases Partitions into sets of sizes 1, 2 and 2 M1 1.1 Considering cases where set sizes are 1, 2, 2 5 × (4C2 ÷ 2) = 5 × 3 = 15 partitions of this type A1 2.1 Explainingwhy there are 15 of these e.g a relevant calculation or list of cases [4] 1 (c) 10 partitions into sets of sizes 1, 1, 1, 2 M1 2.1 Trying to dealwith the cases when there are more than 3 subsets 1 partition into sets of sizes 1, 1, 1, 1, 1 May be implied from answer 51 15 + 25 + 10 + 1 = 51 A1 1.1 51 [2] 1 (d) Number line is split into 6 pieces B1 2.1 6 pieces But there are 8 numbers B1 2.2a Using pigeonhole, or explainingwhy there must be at least one piece Hence result by the pigeonhole principle with two or more numbers [2]Y534/01 Mark Scheme October 2021 Question Answer Marks AO Guidance 2 (a) (i) Next-fit method M1 A1 [2] 1.1 1.1 Bins 1 and 2 correct All correct 2 (a) (ii) First-fit method M1 A1 [2] 1.1 1.1 Bins 1 and 2 correct All correct 2 (a) (iii) 23 18 15 12 8 7 5 First-fit decreasingmethod M1 A1 [2] 1.1 1.1 Ordered list may be seen Bins 1 and 2 correct All correct 2 (b) With ‘online’ lists the items are presented one at a time and the whole list is not known untilthe end. With next-fit and first-fit the items are placed in the order they appear in the list, so these methods can be used ‘offline’ or ‘online’. However, for first-fit decreasingthe whole list needs to be known before it can be sorted, so first-fit decreasing can only be used for an ‘offline’ list. B1 B1 [2] 1.2 2.3 Evidence of understandingwhat ‘online’ means Evidence of realising that ffd cannot be used with an online list (or implied from an appropriate statement aboutnext-fit andfirst-fit) Bin 1 12 Bin 2 23 Bin 3 15 Bin 4 18 8 Bin 5 7 5 Bin 1 12 15 Bin 2 23 7 Bin 3 18 8 Bin 4 5 Bin 5 Bin 1 23 7 Bin 2 18 12 Bin 3 15 8 5 Bin 4 Bin 5Y534/01 Mark Scheme October 2021 Question Answer Marks AO Guidance 2 (c) 88 ÷ 4 = 22, so M is at least 22 But it is not possible to fill 4 bins of capacity22 Since 22 – 18 = 4 which is less than 5 So the 23 would have to be split as 4 and 19 And then there is no 3 to go with the 19 M = 23 is possible e.g. 23 – x and x, 18 + 5, 15 + 8, 12 + 7 Hence, least M is 23 M1 A1 B1 [3] 1.1 2.4 2.2a Identifyingthat M must be at least 22 Showing that M = 22 is not possible Fully correct explanation Showing that M = 23 is possible 3 (a) ABXE B1 [1] 1.1 3 (b) M1 A1 M1 A1 [4] 1.1 1.1 1.1 1.1 AB = 0.6, AC = 1.1 AD = 2.7, AE = 2.8 DF = 0.6, EF = 0.7 BF = 2.7, CF = 2.5 AF = 3.3 or ft from other valu [Show More]

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