RI S1A Permutations and Combinations_Add Prac_Solns

RAFFLES INSTITUTION H2 Mathematics (9758) 2025 Year 6 __________________________________________________________________ Additional Practice Questions for Chapter S1A: Permutations and Combinations Page 1 of 18 Additional Practice Questions for Chapter S1A: Permutations and Combinations (Solutions) 1 From a group of 5 women and 7 men, one of whom is Mr Lee, find how many committees of size 4 can be formed in which (i) there are 2 men and 2 women, (ii) there is at least 1 man and at least 1 woman, (iii) there is at least 1 man and Mr Lee is in the committee. [ (i) 210; (ii) 455; (iii)165] Solution (i) Number of committees with 2 men and 2 women = 7C2  5C2 = 210. (ii) Using the Complement Method, number of committees with at least 1 man and at least 1 woman = (Number of committees without restriction)  (Number of all men committees)  (Number of all women committees) = 12C4  7C4  5C4 = 455 Note : 5 7 10 1 1 2C C C  is not correct. There is double counting. For eg, 1W 1M 2W 2M and 2W 1M 1W 2M are the same selection but counted as 2 different ways in the above computation [You may want to refer to Example 11 of lecture notes] (iii) Since the committee must consist of at least one man and includes Mr Lee, we choose 3 people from the remaining 11 people. Number of ways = 11C3 = 165 2 4 boys, 4 girls and a teacher are to be seated at a round table. How many ways can they be arranged if (i) there is no restriction? (ii) the teacher is to be seated between any 2 girls? (iii) none of the boys are to be seated together? [ (i) 40320; (ii) 8640; (iii) 2880] Solution (i) Number of ways with no restriction = (9 – 1)! = 40320. (ii) The teacher can be seated between any 2 girls in 4C2  2! = 12 ways. Consider the teacher between any 2 girls as one unit: G G B B B B Number of ways = 12  (7 – 1)! = 8640. GTG

Raffles Institution H2 Mathematics 2025 Year 6 __________________________________________________________________________________________ _________________________________________________________________ Additional Practice Questions for Chapter S1A: Permutations and Combinations Page 2 of 18 (iii) We first seat the teacher and 4 girls. Number of ways = (5 – 1)! = 24. Using the Slotting Method, number of ways such that none of the boys are seated together = 24  5P4 = 2880 3 How many 6-digit numbers (i) are even? (ii) begin and end with different digits? [(i) 450000; (ii) 810000] Solution (i) Number of choices for the first digit = 9. Number of choices for each of the 2nd, 3rd, 4th and 5th digit = 10. For the 6-digit number to be even, number of choices for the last digit = 5. Total number of 6-digit numbers that are even = 9 104  5= 450000. (ii) Numbe