ACJC 2025 Permutations and Combinations Summary

Anglo-Chinese Junior College 2025 H2 Mathematics 9758: Permutations & Combinations / Summary / Page 1 of 8 SUMMARY: Permutations and Combinations EXAMPLE 1 Find the number of ways to seat the 10 students in a class of 6 boys and 4 girls Solution (a) in a single row if (i) there is no restriction (ii) the girls sit together (iii) the girls are seated alternately (iv) all the girls are separated (v) one of the girls sits between John and Andy (i) 10! (ii) 7! 4! (iii) 6! 4! 4 or 6C33! 4! 4! (iv) 7C4 4!6! (v) 4C1 8! 2! (b) at a round table if (i) there is no restriction (ii) the girls sit together (iii) the girls are seated alternately (iv) all the girls are separated (v) one of the girls sits between John and Andy (i) (10 1)!− (ii) (7 1)!4!− (iii) 6 3 3!(4 1)!4!C − (iv) 6 4(6 1)! 4!C− (v) 4 1(8 1)!2!C − (c) at a round table where the seats are numbered if (i) there is no restriction (ii) the girls sit together (iii) the girls are seated alternately (iv) all the girls are separated (v) one of the girls sits between John and Andy (i) (10 1)! 10− (ii) (7 1)!4! 10− (iii) 6 3 3! 4!(4 1)! 10C − (iv) 6 4(6 1)! 4! 10C− (v) 4 1 (8 1)!2! 10C − (d) in 2 rows of 5 each, labelled row 1 and row 2, if (i) there is no restriction (ii) the girls sit together (iii) both the girls and boys are seated alternately (i) 10! (ii) 6 1 2!4!5! 2C  (iii) 4! 6! Refer to pg 3 for detailed explanation.

Anglo-Chinese Junior College 2025 H2 Mathematics 9758: Permutations & Combinations / Summary / Page 2 of 8 EXAMPLE 2 Find the number of different arrangements of the word SLEEPLESSNESS, Solution (a) using all thirteen letters if (i) there is no restriction (ii) all “S” are adjacent (iii) all “S” are not next to each other (iv) both the first and last letters are “S” (i) 13! 5!4!2! (ii) 9! 2!4! (iii) 9 5 8! 2!4!C  (iv) 11! 4!3!2! (b) forming 4-letter code words if (i) there is no restriction (ii) both the first and last letters are “S” (i) 5C4 4! (All distinct) + 34 12 4! 2!CC (1 pair same) + 3 2 4! 2! 2!C (2 pairs same) + 24 11 4! 3!CC (3 letters same) + 2 1C (All same) (ii) 5C2 2!+ 3C1 Refer to pg 6 for detailed explanation.

Anglo-Chinese Junior College 2025 H2 Mathematics 9758: Permutations & Combinations / Summary / Page 3 of 8 EXAMPLE 1 Find the number of ways to seat the 10 students in a class of 6 boys and 4 girls Thinking Process Solution (a) in a SINGLE ROW if (i) no restriction The 10 of them can re-arrange among themselves, thus gives 10! 10! (ii) the girls sit together B1 B2 B3 B4 G1G2G3G4 B5 B6 1) Group the girls together in a bubble. 2) The 7 bubbles can re-arrange in different positions, thus 7!. 3) G1 G2 G3 G4 can re-arrange among themselves, thus gives 4!. 7! 4! (iii) the girls are seated alternately Method 1 (List out the possible ways) G1 B2 G2 B3 G3 B4 G4 B1 B5 B6 B1 G1 B2 G2 B3 G3 B4 G4 B5 B6 B1 B5 G1 B2 G2 B3 G3 B4 G4 B6 B1 B5 B6 G1 B2 G2 B3 G3 B4 G4 1) B1 B2 B3 B4 B5 B6 re-arrange among themselves, gives 6!. 2) G1 G2 G3 G4 can re-arrange among themselves, gives 4!. 3) There are 4 ways for the above. 6! 4! 4 Method 2 B1 G1 B2 G2 B3 G3 B4 G4 B5 B6 1) Of the 6 boys, select any 3 boys to be alternating with the girls, which gives 6 3C . The 3 boys selected can re-arrange among themselves, which gives 3!. 2) Ensure that the girls are arranged alternately. 3) Group them as a bubble. 4) The 4 bubbles can re-arrange in different positions, thus 4!. 5) G1 G2 G3 G4 can re-arrange among themselves, gives 4!. 6 3 3! 4! 4!C (iv) all the girls are separated. B1 B2 B3 B4 B5 B6 1) List the remaining group (i.e the boys) with spacing in between. The arrows indicate the space to slot the girls. Note the space for first and last position. 2) Of the 7 arrows, select any 4 arrows to make space for the girls, thus 7 4C . This ensures that the girls are separated. 3) G1 G2 G3 G4 re-arrange among themselves, gives 4!. 4) B1 B2 B3 B4 B5 B6 re-arrange among themselves, gives 6! 7 4 4! 6!C (v) one of the girls sits between John and Andy B1 B2 B3 B4 John G1 Andy G2 G3 G4 1) Of the 4 girls, select a girl to sit between John and Andy, gives 4 1C . Group them together. 2) The 8 bubbles can re-arrange in different positions, thus 8!. 3) John and Andy can swop positions. 4 1 8! 2!C

Anglo-Chinese Junior College 2025 H2 Mathematics 9758: Permutations & Combinations / Summary / Page 4 of 8 (b) at a ROUND TABLE if (i) no restriction The 10 of them can re-arrange among themselves, thus gives ( )10 1 !− 9! (ii) the girls sit together B2 G1 B1 G2 G3 B3 table G4 B4 B6 B5 1) Group the girls together in a bubble. 2) The 7 bubbles can re-arrange in different positions, thus (7– 1)!. 3) G1 G2 G3 G4 can re-arrange among themselves, thus gives 4!. ( )7 1 ! 4!− (iii) the girls are seated alternately G1 B1 B2 G2 B5 table B3 G3 B6 B4 G4 1) Of the 6 boys, select any 3 boys to be alternating with the girls, which gives 6 3C . The 3 boys selected can re-arrange among themselves, which gives 3!. 2) Ensure that the girls are arranged alternately. 3) Group them as a bubble. 4) The 4 bubbles can re-arrange in different positions, thus (4 –1)!. 5) G1 G2 G3 G4 can re-arrange among themselves, gives 4!. ( ) 6 3 3! 4 1 !4!C − (iv) all the girls are separated. B1 B2 B6 table B3 B5 B4 1) List the remaining group (i.e the boys) with spacing in between. The arrows indicate the space to slot the girls. Note the space for first and last position. 2) B1 B2 B3 B4 B5 B6 can re-arrange among themselves, gives (6 –1)!. 3) Of the 6 arrows, select any 4 arrows to make space for the girls, thus 6 4C . This ensures that the girls are separated. 4) G1 G2 G3 G4 can re-arrange among themselves, gives 4!. ( ) 6 46 1 ! 4!C− (v) one of the girls sits between John and Andy B2 G2 B1 G3 B3 table G4 B4 John Andy