1. Permutations and Combinations Solutions 2023 (RVHS)

River Valley High School, Mathematics Department 2023 Permutations and Combinations T1 Permutations and Combinations Solutions 1(i) Number of ways ( ) ( ) 3 girls9 people around a circle 6 boys and 1 blo ck of 3 girls Total number of ways without restriction - number of ways all 3 girls together 9 1 ! 7 1 ! 3! 36000 =  = − − −   = Alternative method Number of ways ( ) 3 2 3 girls choose 2 to be in a group6 boys around a circle 6 2 permute w6 slots permute 2 groups of girls number of ways 2 girls together, 1 separated number of ways all 3 girls separated 5 1 ! C P 2! = + − =  ( ) ithin the group of 2 girls 6 3 6 slots permute 3 girls 6 boys around 5 1 ! P 36000      + −  = (ii) Number of ways ( ) 6 3 The 3 girls6 slots choose 36 remaining people around a circle 6 1 ! 3! 14400 C= −   = (iii) Note: The arrangement must look like this Number of ways without restrictions (9 1)! 40320 =− = B B B B B B G G G

River Valley High School, Mathematics Department 2023 Permutations and Combinations T2 Number of ways for exactly 2 boys between any 2 girls ( ) 6 boys permutate within themselvesArranging 3 girls around a circle 3 1 ! 6! 1440 = −  = Required prob. 1440 40320 1 28 = = Alternative Method: Number of ways without restrictions (9 1)! 40320 =− = Number of ways for exactly 2 boys between any 2 girls ( ) ( ) Arranging 3 girls around a circle 36 4 2 2 2 2 6boys choose 2 to slot into first slot 4 boys choose 2 to slot into second slot each group of boys permute within themselves 3 1 ! C C C 2! 1440 = −        = Required prob. 1440 40320 1 28 = = 2(a)(i) Case 1: 2 red + 2 green, no of ways 58 22 280CC=  = Case 2: 1 red + 3 green, no of ways 58 13 280CC=  = Case 3: 4 green, no of ways 8 4 70C== Total no of ways = 280 + 280 + 70 = 630 (ii) No of ways to select at least 1 of each colour = n(any 4 of 13 balls) – n(4 of 5 red) – n(4 of 8 green) = 13 5 8 4 4 4C C C−− = 640 (b) Case 1: blue trousers, i.e. any 2 of 5 skirts: no of ways 5 2 20P== Case 2: green trousers, i.e. no green skirt: no of ways 4 2 12P== Case 3: yellow trousers, i.e. no yellow skirt: no of ways 4 2 12P==

River Valley High School, Mathematics Department 2023 Permutations and Combinations T3 Total no of ways = 20 + 12 + 12 = 44 3 The number of seating arrangements = 4! = 24 Number of ways = 3!  2! = 12 Number of ways = 4!  2!  6 = 288 4(a) No of ways = 10 9 13 840CC= OR 10 9 31 840CC= (i) Form CLC unit: 3 2 2C  ways Seat 8 units in a round table: (8 1)!− ways Total number of ways = 3 2 2 (8 1)!C   − = 30240 (ii) Total number of ways = 30240 10 302400= (b) Case 1: 4 digit number, ending with 2: No of ways = 2  3  2  1 = 12 Case 2: 4 digit number, does not end with 2: No of ways = 2  (2 + 1)  1 = 6 Case 3: 5 digit number, ending with 2: No of ways = 4! 1 = 24 Case 4: 5 digit number, does not end with 2: No of ways = 4! 2!  2 = 24 Total number of ways = 12 + 6 + 24 + 24 = 66 5(i) No of delegations = 47 23         = 210 (ii) No of groupings = 11 6 3 5 3 3 2!            = 4620 6(i) No of arrangements = 10! = 3628800 (ii) No of arrangements = 5! 6 5!5   = 86400 (iii) No of arrangements = (7-1)! 5!(11) = 950400 7(i) No of words = 5! 4! 22! = 720 (ii) No of words = 6! 2! x 2 = 6! = 720 (iii) No of words = 4! 4! 2! x 2 2 = 288 (iv) No of words = 54! 4! 42! 2   = 720 8(a) No of arrangements = 6! 2! = 360

River Valley High School, Mathematics Department 2023 Permutations and Combinations T4 (b) No of selections = 4 1   x 2 = 8 9(i) No of 4-digit numbers = 3 x 3! = 18 (ii) No of 4-digit numbers = 2! x 5 = 10 10(i) Total number of possible results = (ii) Number of ways of obtaining ‘Windfall’ = 3 (iii) No. of ways to obtain a success = 5! 5! 2 3 1203!2! 3!   +  = Total number of ways 120 Reqd no. of ways = 10 7 4 3 3 4 3! 126002!CCC   = 11(a)(i) Required number of 8-letter code-words = 85 = 390625 (ii) Required number of 8-letter code-words = ( )3! 2 ! 8 = 5040 (iii) First, consider that each letter occurs once, then we add 3 more letters (from A, B, C, D, E) to form the code-words. Case 1: 3 identical letters No. of ways = 5 Case 2: 3 different letters No. of ways = 3 5C =10 Case 3: 2 identical letters and 1 different letter No. of ways = 22 5 C = 20 Required number of 8-letter code-words = 5 + 10 + 20 =35 (b) Case 1: 2, 2, 5 people No. of ways = 9072! 4 ! 1 ! 1! 2 2 7 2 9 = CC Case 2: 2, 3, 4 people No. of ways = 15120! 3 ! 2 ! 13 7 2 9 = CC Case 3: 3, 3, 3 people No. of ways = ( ) 2240! 2! 3 33 6 3 9 = CC Total no. of ways = 26432 53 243= 243=− 123=

River Valley High School, Mathematics Department 2023 Permutations and Combinations T5 12(a)(i) No. of arrangements 4! 5! 14402!=  = (ii) No. of arrangements 3 24! 144P=  = (b) No. of arrangements (10 1)! 2! 11 7983360= −   = Case 1: Mr Lin and Mr Tan at Circular Table No. of arrangements = 9 4(5 1)! 2 5! 725760C −   = Case 2: Mr Lin and Mr Tan at Linear Table No. of arrangements = 9 3(6 1)! 2 4! 483840C −   = Total number of arrangements = 1209600 13(a) Case 1: 4 and 5 in the second and fourth positions 2 3! 12= Case 2: 3 and 5 in the second and fourth positions 2 2 4= Total number 12 4 16= + = (b) Number of ways 4 3 4 3 4 3 . . .3! 2041 3 2 2 3 1            =  + + =                       Alternative: 7 4.3! 3! 204C −= 14(i) 65 24 75CC= ways (ii) Number of ways if at least one of the sisters are included = number of ways without restriction – number of ways if none of the sisters is included = 11 8 66 434CC−= Or 3 8 3 8 3 8 1 5 2 4 3 3 434C C C C C C +  +  = (iii) Select a man to be between the 2 sisters and group the 3 of them as one unit and arrange 4 units round a table 3 1 3! 2 36C   = (iv) First arrange the other 4 persons round the table. There are 4 ways to insert the sisters. 3! 4 24= or 4 2 2! 2! 24C   = 15 For distinct gifts, 65 ways Now considering the distinct gifts, Case 1: 3 person get 1 gift

River Valley High School, Mathematics Department 2023 Permutations and Combinations T6 No of ways = 5 6 3 1562505C  = Case 2: 1 person get 1 gift, another person gets 2 gifts No of ways = ( ) 5 6 2 2 3125005C  = Case 3: 1 person get 3 gifts No of ways = 5 6 1 5 78125C = Total number of ways 156250 312500 78125 546875= + + = Alternative Stage 1: Distribute 6 distinct gifts among 5 people No of ways = 65 Stage 2: Distribute 3 identical gifts among 5 people Case 1: 3 person get 1 gift No of ways = 5 3C = 10 Case 2: 1 person get 1 gift, another person gets 2 gifts No of ways = ( ) 5 2 2C = 20 Case 3: 1 person get 3 gifts No of ways = 5 1C = 5 Total number of ways = (10+20+5)56 = 546875 16(i) Number of ways 8! 40320== (ii) Number of ways 4!×4!×4 2304 = = (iii) Number of ways 10 10 88 10!×8! or or 2! 1814400 CP= = (iv) Number of ways 48 26 48 26 ×2! 6! or 241920 CC PP =    = 4 = ways to arrange the boys and girls on each side 4!×4! = ways the groups of 4 boys & 4 girls can arrange themselves 10 8C = ways to choose any 8 out of the 10 seats to sit the 4 boys and 4 girls 8! = ways to arrange the boys and girls 10! arrange 10 objects in a row with 2 identical 2! objects (empty seats) = 4 2×