RI S1A Permutations and Combinations tutorial (Solns)

RAFFLES INSTITUTION H2 Mathematics (9758) 2025 Year 6 __________________________________________ Tutorial S1A: Permutations and Combinations Page 1 of 17 Tutorial S1A: Permutations and Combinations Section A (Discussion Questions) 1 9 coloured balls are arranged in a row. Find the number of ways this can be done if (i) all the balls are of different colours, (ii) there are 4 red balls, 3 green balls and 2 yellow balls, (iii) there are 4 red balls, 3 green balls and 2 yellow balls, and the arrangement is symmetrical. [(i) 362880; (ii) 1260; (iii) 12 ] Solution (i) Number of ways = 9! = 362880 9 coloured balls, 4 R, 3 G, 2 Y (ii) Number of ways = 9! 4!3!2! =1260 Alternative method: Out of the 9 positions, choose 4 for the 4 red balls, out of the remaining 5 positions, choose 3 for the 3 green balls : 95 43CC (iii) there are 4 red balls, 3 green balls and 2 yellow balls, and the arrangement is symmetrical. _____ _____ _____ _____ G _____ _____ _____ _____ 2R , 1G, 1Y 2R , 1G, 1Y Since arrangement is symmetrical, middle ball must be green. The first 4 balls must consist 2 red, 1 green and 1 yellow ball. Number of ways = 4! 2! =12 Alternative method: 42 21CC Choose the slots instead.

Raffles Institution H2 Mathematics 2025 Year 6 _____________________________________________________________________________________________ ______________________________________ Tutorial S1A: Permutations and Combinations Page 2 of 17 2 How many different three figure numbers between 100 and 999 inclusive, contain two and only two consecutive identical figures? [162] Solution Case 1: First 2 digits are identical. Choices for 1st two digits = 9 (1-9, 0 excluded) Choices for 3rd digit = 9 (any digit from 0-9 that has not been used for 1st digit) Number of ways = 9 x 9 = 81 Case 2: Last 2 digits are identical Choices for 1st digit = 9 (1-9, 0 excluded) Choices for 2nd and 3rd digits = 9 (any digit from 0-9 that has not been used for 1st digit) Number of ways = 9 x 9 = 81 Total number of three figure numbers = 81 + 81 = 162 3 NJC Prelim 9233/2005/01/Q7 (modified) (a) A resort hotel has 3 single rooms, 5 double rooms and 4 family rooms. On a particular week, 2 individuals book a single room each, 3 couples book a double room each and 3 families book a family room each. Given th at all the rooms are available for that week and each room has a distinct design theme, find the number of different possible bookings of rooms amongst the guests. (b) A family of 6 adults and 3 children are to be seated at a round table. Find the number of possible arrangements such that (i) all the 3 children sit together, (ii) none of the children sit together. [ (a) 8640; (b)(i) 4320; (ii) 14400] Solution (a) Rooms : 3 single, 5 double, 4 family People : 2 individual, 3 couples, 3 families Number of arrangements for singles = 3 2P Number of arrangements for couples = 5 3P Number of arrangements for families = 4 3P Total number of arrangements = 354 233P PP =8640 (b) 6 adults, 3 children, seated at a round table (i) all the 3 children sit together C C C A A A A A A - 7 units

Raffles Institution H2 Mathematics 2025 Year 6 _____________________________________________________________________________________________ ______________________________________ Tutorial S1A: Permutations and Combinations Page 3 of 17 Number of ways to arrange 7 units at a round table = (7  1)! Number of arrangements within the “children unit” = 3! Total number of seating arrangements = (7  1)!  3! = 4320 (ii) none of the children sit together Slotting Method: Number of arrangements to first seat the 6 adults = (6  1)! = 120 Number of arrangements to slot in the 3 children in the 6 empty spaces between the adults = 6 3P Total number of seating arrangements = (6  1)!  6P3 = 14400 4 (a) Twelve people are to make a journey in 3 cars, with 4 people in each car. Each car is driven by its owner. Find the number of ways in which the remaining 9 people may be allocated to the cars. [The arrangement of people within each car is not relevant.] (b) Nine people are to be divided into 4 groups. Find the number of ways this can be done if 1 group consists of 3 people and the other 3 groups consist of 2 people each. [ (a) 1680; (b) 1260] Solution (a) Each group is distinct as there is a unique car owner. Number of ways = 963 333CCC = 1680 OR 9! 3!3!3! Arrange all the people in a line. First 3 people to Car 1, next 3 to Car 2 and last 3 to Car 3. However arrangement is not needed within each car, so divide by 3!3!3!. (b) The 3 groups of 2 are identical, and there is a need to divide by 3! Number of ways = 9753 2223 3! CCCC = 1260 5 A student has 8 different Literature books, 3 different History books and 3 different Geography books. How many ways can 5 books be selected if the selection must consist of at least one book of each subject? [The order of selection is not relevant]. [1128] Solution Lit (8) Histor y (3) Geo g (3) No. of wa ys 3 1 1 83 3 311CCC = 504 2 2 1 2( 833 221CCC ) = 504 2 1 2 1 3 1 833 131CCC = 24 1 2 2 83 3 122CCC = 72 1 1 3 833 131CCC =24 Summing up the above, total number of ways = 1128 A A A A A A

Raffles Institution H2 Mathematics 2025 Year 6 _____________________________________________________________________________________________ ______________________________________ Tutorial S1A: Permutations and Combinations Page 4 of 17 Alternative 1: Complement method Number of choices, without restriction = 14 5C = 2002 Complement cases: Case 1: All 5 books chosen are History or Geog Books ONLY Number of choices = 6 5C Case 2: All 5 books chosen are Literature or History Books ONLY Number of choices = 11 5C Case 3: All 5 books chosen are Literature or Geog Books ONLY Number of choices = 11 5C Note that in cases 2 and 3, both include the case where only literature books are chosen Number of choices, when only literature books are chosen = 8 5C =56 Hence, number of choices =  14 6 11 11 8 5 555 5C CCC C = 1128 Note : 8331 1 111 2CCC C is not correct. There is double counting. You can make reference to Eg 11 of lecture notes For eg, 1L 1H 1G 2G 3G 1L 1H 2G 1G 3G

Raffles Institution H2 Mathematics 2025 Year 6 _____________________________________________________________________________________________ ______________________________________ Tutorial S1A: Permutations and Combinations Page 5 of 17 6 9740/2009/02/Q8 Find the number of ways in which the letters of the word ELEVATED can be arranged if (i) there are no restrictions, [1] (ii) T and D must not be next to one another, [2] (iii) consonants (L, V, T, D) and vowels (E, A) must alternate, [3] (iv) between any two Es there must be at least 2 other letters. [3] [(i) 6720 ; (ii) 5040; (iii) 192 (iv) 480] Solution 3 E’s, L, V, A, T, D ------ 8 letters (i) Number of arrangements without restriction = 8! 3! = 6720 (ii) T and D must not be next to one another Method 1 : Slotting method __ E __ E __ E __ L __ V __ A __ Number of arrangements where T and D are not next to one another = 6! 3! 7 2 2!C = 5040 or 6! 3! 7P2 Method 2 : Complement method TD E E E L V A Number of arrangements where T and D are together 7! 2! 16803! Number of arrangements where T and D are not next to one another 6720 1680 5040 (iii) Consonants (L, V, T, D) and vowels (