SAJC 2021 JC1 H1 Maths Final Exam concepts

Page 1 2021 JC1 H1 Maths FE level up concepts Holiday REVISION  Please present all your answers on this copy as it will be submitted to your tutor when Term 1 starts in 2022.  If in doubt, please read your lecture notes, if still further in doubt, put a note under ‘Learning Point’ so that you remember to clarify your doubts with your tutor. Where possible, answers should be left as fraction. Question Solutions Learning Point 1. Permutation and Combination (Arrangement) (i) Three friends, Andy, Ben and Charlie, attend a graduation ceremony with their parents. How many ways can the nine people be seated in a row if each family must be seated together? Ans: 1296 Arrange within family = 3! Arrange 3 families = 3! 3! 3! Total arrangement = 3! 3! 3! 3! 1296    (ii) How many ways can the nine people be seated in a row if each set of parents must be seated together? Ans: 5760 Arrange within each set of parents = 2! Arrange 3 sets of parents and 3 friends = 6! Total arrangement =   3 2! 6! 5760 (iii) How many ways can the nine people be seated in a row if the 3 friends must be seated together? Ans: 30240 Arrange within the 3 friends (1 unit) = 3! Arrange 1 unit and 6 parents = 7! Total arrangement = 7! 3! 30240

Page 2 (iv) How many ways can the nine people be seated in a row if the 3 friends wants to be separated? Ans:151200 Arrange all 6 parents first (no restriction) = 6! SLOT the 3 friends into the 6 parents = 7 3P Total arrangement = 7 3 6! 151200P  (v) How many ways can the nine people be seated in a row if each set of parents must be seated together AND the 3 friends wants to be separated? Ans:1152 Arrange within each set of parents =   3 2! Arrange 3 sets of parents = 3! SLOT the 3 friends into the 3 sets of parents = 4 3P (Think carefully here!) Total arrangement =   3 4 32! 3! 1152P   (vi) How many ways can the nine people be seated in a row if Andy must take the first seat, Ben must take the last seat and each set of parents must be seated together? Ans:192 Arrange within each set of parents =   3 2! Arrange Charlie and 3 sets of parents = 4! Total arrangement =   3 2! 4! 192

Page 3 2. Permutation and Combination (Combination) In a college student committee, there are 4 Administrators, 5 Liaison officers and 6 Group Leaders. They are to form an Ex-Co that consists of a President, a Vice-President and a Secretary. (i) How many selections are there to form the Ex-Co? Ans 2730 Total selection (no restriction) = 15 3P 2730 (ii) How many selections are there to form the Ex-Co if a President must be from the Administrators, a Vice- President must be from the Liaison Officers and a Secretary must be from the Group Leaders? Ans : 120 To select a President from Administrators = 4 1C To select a Vice President from the Liaison Officers =