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) Number of choices for the first digit = 9. Since the first and last digits are different, number of choices for the last digit = (10  1) = 9. Number of choices for each of the 2nd, 3rd, 4th and 5th digit = 10. Total number of 6-digit numbers that begin and end with different digits = 9  9  104 = 810000 4 A rectangular table has 7 secured seats, 4 being on one side facing the window and 3 being on the opposite side. In how many ways can 7 people be seated at the table (i) if 3 people, X and Y and Z must sit on the side facing the window? (ii) if 2 people, P and Q must sit on opposite sides? [ (i) 576; (ii) 2880] Solution WINDOW G G G G T

Raffles Institution H2 Mathematics 2025 Year 6 __________________________________________________________________________________________ _________________________________________________________________ Additional Practice Questions for Chapter S1A: Permutations and Combinations Page 3 of 18 (i) if 3 people, X and Y and Z must sit on the side facing the window We choose one more person to sit on the same side as X, Y, Z Number of choices = 4 1C = 4 The people in each row can be permuted within the row. Total number of arrangements = 4 x 4! x 3! = 576 Altenative 1: Number of ways to arrange 3 out of the remaining 4 people to sit on the side near the window = 4P3 Number of ways to arrange X, Y, Z and the 4th person = 4! Total number of arrangements = 4P3  4! = 576 Alternative 2: Number of ways to choose 3 seats for X, Y, Z , and arrange them = 4 3 3!C  Number of ways to arrange the remaining 4 people = 4! Total number of arrangements = 4 3 3!C   4! = 576 (ii) if 2 people, P and Q must sit on opposite sides Case 1 : P is on the side with 4 seats, Q on the side with 3 seats. Just like in (i), we choose 3 people to sit with P (or 2 people to sit with Q), and permute within the rows. Number of choices 5 3 4! 3!C   Case 2 : Q is on the side with 4 seats and P is on the side with 3 seats. Number of choices is as above. Hence, total number of choices 5 3 4! 3! 2 2880C     Alternative method: Case 1: P sits on the side with 4 seats Number of ways to choose a seat for P = 4 1C Number of ways to choose a seat for Q = 3 1C Number of ways to arrange the remaining 5 people = 5! Number of ways 4 3 1 1 5!C C   Case 2: Q sits on the side with 4 seats Number of ways is as above. Total number of ways 4 3 1 1 5! 2 2880C C    

Raffles Institution H2 Mathematics 2025 Year 6 __________________________________________________________________________________________ _________________________________________________________________ Additional Practice Questions for Chapter S1A: Permutations and Combinations Page 4 of 18 5 9205/1989/01/Q19(b) A school is asked to send a delegation of six pupils selected from six badminton players, six tennis players and five squash players. No pupil plays more than one game. The delegation is to consist of at least one, and not more than three, players drawn from each sport. Giving full details of your working, find the number of ways in which the delegation can be selected. [9450] Solution We use Systematic Listing and consider 7 cases. Number of Badminton players(6) Number of Tennis players (6) Number of Squash players (5) Number of ways to select the 6 players 1 2 3 6C1  6C2  5C3 = 900 1 3 2 6C1  6C3  5C2 = 1200 2 1 3 6C2  6C1  5C3 = 900 2 3 1 6C2  6C3  5C1 = 1500 3 1 2 6C3  6C1  5C2 = 1200 3 2 1 6C3  6C2  5C1 = 1500 2 2 2 6C2  6C2  5C2 = 2250 Total number of ways = 900 + 1200 + 900 + 1500 + 1200 + 1500 + 2250 = 9450. 6 AJC Prelim 9233/2003/01/Q7 Find the number of distinct arrangements of the letters of the word 'THERMOMETER' (i) if at least 2 'E's are together, (ii) which must start and end with 'T' or 'R'. [(i) 408240; (ii) 90720] Solution (i) There are 11 letters, with 3 'E's, 2 'T's, 2 'M's and 2 'R's, 1 ‘O’, 1 “H” Use the Complement Method: Number of ways without restriction = 11! 3!2!2!2!= 831600.  T  H  R  M  O  M  T  R  Using the Slotting Method: Number of ways to arrange the 2T, 2M, 2R, 1O, 1H = 8! 2!2!2!= 5040 The 3 ‘E’s can occupy 3 out of the remaining 9 slots in 9C3 = 84 ways. Number of ways such that the 3 'E's are always separated = 5040  84 = 423360. Therefore, using the Complement Method, number of ways such that at least 2 'E's are together = 831600  423360 = 408240.

Raffles Institution H2 Mathematics 2025 Year 6 __________________________________________________________________________________________ _________________________________________________________________ Additional Practice Questions for Chapter S1A: Permutations and Combinations Page 5 of