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Combinatorics, EXT1 A1 2025 HSC 13b

Eight guests are to be seated at a round table. If two of these guests refuse to sit next to each other, how many seating arrangements are possible?   (2 marks)

Show Answers Only

\(\text{Total combinations}\ = 3600\)

Show Worked Solution

\(\text{Strategy 1}\)

\(\text{If no restrictions:}\)

\(\text{Total combinations\(=7!\)}\)

\(\text{If two people must sit together:}\)

\(\text{Combinations\(=6!2!\)}\)

\(\text{If two people refuse to sit together:}\)

\(\text{Combinations\(=7!-6!2!=3600\)}\)
 

\(\text{Strategy 2}\)

\(\text{Sit one of the refusers in any seat.}\)

\(\text{Possible seats for other refuser = 5}\)

\(\text{Combinations for other 6 people}\ =6!\)

\(\text{Total combinations}\ = 5 \times 6! = 3600\)

Combinatorics, EXT1 A1 EQ-Bank 1 MC

A table tennis club consists of 6 males and 5 females.

How many committees of 4 players can be chosen that contain no more than 2 females?

  1. 250
  2. 265
  3. 305
  4. 330
Show Answers Only

\(\Rightarrow B\)

Show Worked Solution

\(\text{Combinations (0 females)}={ }^6 C_4=15\)

\(\text{Combinations (1 female)}={ }^5 C_1 \times{ }^6 C_3=100\)

\(\text{Combinations (2 females)}={ }^5 C_2 \times{ }^6 C_2=150\)

\(\text{Total combinations }=15+100+150=265\)

\(\Rightarrow B\)

Combinatorics, EXT1 A1 EQ-Bank 3

Four girls and four boys are to be seated around a circular table. In how many ways can the eight children be seated if:

  1. there are no restrictions?   (1 mark)

  2. the two tallest boys must not be seated next to each other?   (1 mark)

  3. the two youngest children sit together?   (1 mark)

Show Answers Only

a.   \(\text{Combinations (no restriction)}\ = 7! \)

b.   \(\text{Total combinations}\ = 5 \times 6! \)

c.   \(\text{Total combinations}\ = 2 \times 6! \)

Show Worked Solution

a.   \(\text{Fix one child in a seat (strategy for circle combinations):}\)

\(\text{Combinations (no restriction)}\ = 7! \)
 

b.   \(\text{Fix the tallest boy in a seat:}\)

\(\text{Possible seats for 2nd tallest boy}\ =5\)

\(\text{Combinations for other 6 children}\ = 6! \)

\(\text{Total combinations}\ = 5 \times 6! \)
 

c.   \(\text{Fix the youngest in a seat:}\)

\(\text{Possible seats for 2nd youngest}\ =2\)

\(\text{Combinations for other 6 children}\ = 6! \)

\(\text{Total combinations}\ = 2 \times 6! \)

Combinatorics, EXT1 A1 2023 HSC 10 MC

A group with 5 students and 3 teachers is to be arranged in a circle.

In how many ways can this be done if no more than 2 students can sit together?

  1. \(4 ! \times 3!\)
  2. \(5 ! \times 3!\)
  3. \(2 ! \times 5 ! \times 3!\)
  4. \(2 ! \times 2 ! \times 2 ! \times 3!\)
Show Answers Only

\(B\)

Show Worked Solution

\(\text{Fix 1st teacher in a seat}\)

\(\text{Split remaining 5 students into 3 groups (2 × 2 students and 1 × 1 student)}\)
 

\(\text{Combinations of other teachers = 2! }\)

\(\text{Combinations of students within groups = 5! }\)

\(\text{Combinations of student groups between teachers = 3 }\)

\(\therefore\ \text{Total combinations}\ = 2! \times 5! \times 3 = 3! \times 5! \)

\(\Rightarrow B\)

♦♦♦ Mean mark 13%.

Combinatorics, EXT1 A1 2023 HSC 11b

In how many different ways can all the letters of the word CONDOBOLIN be arranged in a line?  (2 marks)

Show Answers Only

\(302\ 400\)

Show Worked Solution

\(\text{CONDOBOLIN → }\ 3 \times \text{O}, 2 \times \text{N}, 1 \times \text{C, D, B, L, I} \)

\(\text{Combinations}\) \(=\dfrac{10!}{3! \times 2!} \)  
  \(=302\ 400\)  

Combinatorics, EXT1 A1 2022 HSC 7 MC

The diagram shows triangle `A B C` with points chosen on each of the sides. On side `A B`, 3 points are chosen. On side `A C`, 4 points are chosen. On side `B C`, 5 points are chosen.
 


 

How many triangles can be formed using the chosen points as vertices?

  1. 60
  2. 145
  3. 205
  4. 220
Show Answers Only

`C`

Show Worked Solution

`text{1 point taken from each side:}`

`text{Triangles} = 3 xx 4xx5=60`
 

`text{2 points taken from one side:}`

`text{Triangles}` `=((3),(2))((9),(1))+((4),(2))((8),(1))+((5),(2))((7),(1))`  
  `=145`  

 
`:.\ text{Total triangles} =60+145=205`

`=>C`


♦♦ Mean mark 37%.

Combinatorics, EXT1 A1 SM-Bank 21

Eight points `P_1, P_2, ..., P_8`, are arranged in order around a circle, as shown below.
 

  1. How many triangles can be drawn using these points as vertices?   (1 mark)

  2. How many pairs of triangles can be drawn, where the vertices of each triangle are distinct points?   (2 marks)

Show Answers Only
  1. `56`
  2. `280`
Show Worked Solution
i.    `text{Total triangles}` `= \ ^8 C_3`
    `= 56`
COMMENT: In part (ii), divide by 2 to account for duplicate pairs.

 

ii.   `text{Total pairs}` `= (\ ^8 C_3 xx \ ^5 C_3)/{2}`
    `= 280`

Combinatorics, EXT1 A1 SM-Bank 20

How many rectangles, including all squares, can be found in the 4 × 5 grid below, in total?   (2 marks)
 

Show Answers Only

`150`

Show Worked Solution

`text{Consider the 5 horizontal lines}`

`text{Height combinations of rectangle} =\ ^5C_2`

`text{Consider the 6 vertical lines}`

`text{Width combinations of rectangle} =\ ^6C_2`
  

`:. \ text{Total rectangles in grid}`

`= \ ^5C_2 xx \ ^6C_2`

`= 150`

Combinatorics, EXT1 A1 2020 HSC 8 MC

Out of 10 contestants, six are to be selected for the final round of a competition. Four of those six will be placed 1st, 2nd, 3rd and 4th.

In how many ways can this process be carried out?

  1. `(10!)/(6!4!)`
  2. `(10!)/(6!)`
  3. `(10!)/(4!2!)`
  4. `(10!)/(4!4!)`
Show Answers Only

`C`

Show Worked Solution
`text(Combinations)` `= \ ^10 C_6 xx \ ^6P_4`
  `= \ ^10 C_6 xx 6 xx 5 xx 4 xx 3`
  `= (10!)/(6!4!) xx (6!)/(2!)`
  `= (10!)/(4!2!)`

  
`=>C`

Combinatorics, EXT1 A1 SM-Bank 6

  1. In how many ways can the numbers 9, 8, 7, 6, 5, 4 be arranged around a circle?   (1 mark)

  2. How many of these arrangements have at least two odd numbers together?    (2 marks)

Show Answers Only
  1. `120`
  2. `108`
Show Worked Solution

i.    `text{Fix 9 (or any odd number) on the circle}.`

`text(Arrangements) \ = 5 ! = \ 120`
  

ii.     `text(Fix) \ 9 \ text(on circle).`

  `text(Consider arrangements with no odd numbers together):`
 


 

`text{Combinations (clockwise from top)}`

`= 1 × 3 × 2 × 2 × 1 × 1`

`= 12`
 

`:. \ text(Arrangements with at least 2 odds together)`

`= 120 – 12`

`= 108`

Combinatorics, EXT1 A1 SM-Bank 5

  1. In how many ways can the letters of COOKBOOK be arranged in a line?  (1 mark)

  2. What is the probability that a random rearrangement of the letters has four O's together?  (2 marks)

Show Answers Only
  1. `840`
  2. `1/14`
Show Worked Solution

i.   `text(Combinations) = (8!)/(4!2!) = 840`

 

ii.   `text(Treat four O’s as one letter)`

`text(Combinations) = 5! = 120`

`text(Adjusting for 2 Ks:)`

`text(Combinations) = (5!)/(2!) = 60`

`:. P(text(4 O’s together))`

`= 60/840`

`= 1/14`

Combinatorics, EXT1 A1 EQ-Bank 4

How many numbers greater than 6000 can be formed with the digits 1, 4, 5, 7, 8 if no digit is repeated.  (2 marks)

Show Answers Only

`168`

Show Worked Solution

`text(4 digit numbers:)`

`text(Must start with 7 or 8)`

`text(Combinations (4 digits)) = 2 xx\ ^4P_3 = 2 xx 4 xx 3 xx 2 xx 1= 48`
 

`text(5 digit numbers:)`

`text(Combinations) = 5! = 120`

`:.\ text(Total combinations = 168)`

Combinatorics, EXT1 A1 2019 HSC 8 MC

In how many ways can all the letters of the word PARALLEL be placed in a line with the three Ls together?

  1. `(6!)/(2!)`
  2. `(6!)/(2!3!)`
  3. `(8!)/(2!)`
  4. `(8!)/(2!3!)`
Show Answers Only

`A`

Show Worked Solution

`text(Treat 3 L’s as one letter).`

♦ Mean mark 47%.

`text(Combinations of 6 different letters = 6!)`

`text(Adjusting for 2 A’s:)`

`text(Combinations) = (6!)/(2!)`

`=>\ text(A)`

Combinatorics, EXT1′ S1 2019 HSC 10 MC

An access code consists of 4 digits chosen from the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. The code will only work if the digits are entered in the correct order.

Some access codes contain exactly two different digits, for example 3377 or 5155.

How many such access codes can be made using exactly two different digits?

  1. 630
  2. 900
  3. 1080
  4. 2160
Show Answers Only

`A`

Show Worked Solution

`text(Code has 1x first digit, 3x other digit:)`

`text(First digit = 10 choices)`

`text(Position of first digit = 4 choices)`

`text(Other digit = 9 choices)`

`text(Combinations)` `= 10 xx 4 xx 9`
  `= 360`

 
`text{Code has 2 × each digit  (say}\ X and Y\text{):}`

`text(Combinations of 2 digits) = \ ^10 C_2`

`text(If)\ X\ text(in position 1) => 3\ text(combinations of last 3 digits)`

`text(If)\ Y\ text(in position 1) => 3\ text(combinations of last 3 digits)`

`text(Combinations)` `= \ ^10 C_2 xx 6`
  `= 270`

 
`:.\ text(Total access codes) = 360 + 270 = 630`

`=>   A`

Combinatorics, EXT1 A1 2018 HSC 8 MC

Six men and six women are to be seated at a round table.

In how many different ways can they be seated if men and women alternate?

A.     `5!\ 5!`

B.     `5!\ 6!`

C.     `2!\ 5!\ 5!`

D.     `2!\ 5!\ 6!`

Show Answers Only

`B`

Show Worked Solution

`text(Position the 1st man in any seat.)`

♦ Mean mark 42%.

`text(The remaining 5 men can be positioned in 5! ways).`

`text(The 6 women can be positioned in the alternate seats)`

`text(in 6! ways.)`
 

`:.\ text(Total seating arrangements)\ = 5! xx 6!`
 

`⇒  B`

Combinatorics, EXT1 A1 2017 HSC 10 MC

Three squares are chosen at random from the 3 × 3 grid below, and a cross is placed in each chosen square.
 


 

What is the probability that all three crosses lie in the same row, column or diagonal?

A.     `1/28`

B.     `2/21`

C.     `1/3`

D.     `8/9`

Show Answers Only

`B`

Show Worked Solution
`P` `= text(favourable events)/text(total possible events)`
  `= (3 + 3 + 2)/(\ ^9C_3)`
  `= 2/21`

 
`=>B`

Combinatorics, EXT1 A1 2016 HSC 8 MC

A team of 11 students is to be formed from a group of 18 students. Among the 18 students are 3 students who are left-handed.

What is the number of possible teams containing at least 1 student who is left-handed?

  1. `19\ 448`
  2. `30\ 459`
  3. `31\ 824`
  4. `58\ 344`
Show Answers Only

`B`

Show Worked Solution

`text(Teams with at least 1 left-hander)`

`=\ ^18C_11 -\ ^15C_11`

`= 30\ 459`
 

`=>   B`

Combinatorics, EXT1′ A1 2007 HSC 5a

A bag contains 12 red marbles and 12 yellow marbles. Six marbles are selected at random without replacement.

  1. Calculate the probability that exactly three of the selected marbles are red. Give your answer correct to two decimal places.   (1 mark)

  2. Hence, or otherwise, calculate the probability that more than three of the selected marbles are red. Give your answer correct to two decimal places.   (2 marks)

Show Answers Only
  1. `0.36`
  2. `0.32`
Show Worked Solution

i.   `\ ^12C_3= text(# Ways of selecting 3 R or Y from 12.)`

`\ ^24C_6=text(# Ways of selecting 6 from 24.)`

`P text{(exactly 3R)}` `=(\ ^12C_3 xx \ ^12C_3)/(\ ^24C_6)`
  `=(220 xx 220)/(134\ 596)`
  `=0.36\ \ text{(to 2 d.p.)}`

 

ii.   `text(Solution 1)`

`text(S)text(ince)\ \ P text{(> 3 Red)` `=Ptext{(< 3 Red)}`
 `P text{(> 3 Red)` `=1/2[1-Ptext{(exactly 3R)}]`
  `=1/2(1-0.36)`
  `=0.32`

 

`text(Solution 2)`

`P (> 3R)` `=P (4R) + P (5R) + P (6R)`
  `=(\ ^12C_4  \ ^12C_2 + \ ^12C_5  \ ^12C_1 + \ ^12C_6 xx \ ^12C_0)/(\ ^24C_6)`
  `=(43\ 098)/(134\ 596)`
  `=0.32\ \ text{(to 2 d.p.)}`

Combinatorics, EXT1 A1 2007 HSC 5b

Mr and Mrs Roberts and their four children go to the theatre. They are randomly allocated six adjacent seats in a single row.

What is the probability that the four children are allocated seats next to each other?  (2 marks)

Show Answers Only

`1/5`

Show Worked Solution

`text(Total combinations possible)`

`= 6! = 720`
 

`text(Consider the children as 4 individuals in a group)`

`=>\ text(Combinations)\ = 4! = 24`
 

`text(Consider the two parents and a group of 4)`

`text(children as 3 elements.)`

`=>\ text(Combinations)\ = 3! = 6`
 

`:.\ text{P(children all sit next to each other)}`

`= (6 xx 24)/720`

`= 1/5`

Combinatorics, EXT1 A1 2004 HSC 4c

Katie is one of ten members of a social club. Each week one member is selected at random to win a prize.

  1. What is the probability that in the first 7 weeks Katie will win at least 1 prize?  (1 mark)

  2. Show that in the first 20 weeks Katie has a greater chance of winning exactly 2 prizes than of winning exactly 1 prize.  (2 marks)

  3. For how many weeks must Katie participate in the prize drawing so that she has a greater chance of winning exactly 3 prizes than of winning exactly 2 prizes?  (2 marks)

Show Answers Only
  1. `0.52`
  2. `text(See Worked Solutions)`
  3. `text(30 weeks)`
Show Worked Solution

i.  `Ptext{(wins at least 1 prize)}`

`= 1 − Ptext{(wins no prize)}`

`= 1 − (9/10)^7`

`= 0.5217…`

`= 0.52\ \ \ text{(2 d.p.)}`
 

ii.  `text(In 1st 20 weeks,)`

`Ptext{(winning exactly 1 prize)}`

`=\ ^(20)C_1 · (1/10) · (9/10)^19`

`= 0.2701…`
 

`Ptext{(winning exactly 2 prizes)}`

`=\ ^(20)C_2 · (1/10)^2 · (9/10)^18`

`= 0.2851…`
 

`:.\ text(Katie has a greater chance of winning)`

`text(exactly 2 prizes.)`

 

iii. `Ptext{(winning exactly 3 prizes)}`

`=\ ^nC_3 · (1/10)^3 · (9/10)^(n − 3)`

`Ptext{(winning exactly 2 prizes)}`

`=\ ^nC_2 · (1/10)^2 · (9/10)^(n − 2)`
 

`text(If greater chance of winning exactly 3 than exactly 2:)`

`\ ^nC_3 · (1/10)^3 · (9/10)^(n − 3)` `>\ ^nC_2 · (1/10)^2 · (9/10)^(n − 2)`
`(n!)/(3!(n − 3)) · 1/10` `> (n!)/(2!(n − 2)!) · 9/10`
`(2!(n − 2)!)/(3!(n − 3)!)` `> 9`
`(n − 2)/3` `> 9`
`n − 2` `> 27`
`n` `> 29`

 
`:.\ text(Katie must participate for 30 weeks.)`

Combinatorics, EXT1 A1 2004 HSC 2e

A four-person team is to be chosen at random from nine women and seven men.

  1. In how many ways can this team be chosen?  (1 mark)

  2. What is the probability that the team will consist of four women?  (1 mark)

Show Answers Only
  1. `1820`
  2. `9/130`
Show Worked Solution

i.   `text(# Team combinations)`

`=\ ^(16)C_4`

`= (16!)/((16 − 4)!\ 4!)`

`= 1820`
 

 ii.  `text{P(4 women)}`

`= (\ ^9C_4)/1820`

`= 126/1820`

`= 9/130`

Combinatorics, EXT1 A1 2006 HSC 3c

Sophie has five coloured blocks: one red, one blue, one green, one yellow and one white. She stacks two, three, four or five blocks on top of one another to form a vertical tower.

  1. How many different towers are there that she could form that are three blocks high?  (1 mark)

  2. How many different towers can she form in total?  (2 marks)

Show Answers Only
  1. `60`
  2. `320`
Show Worked Solution
i.   `text(Towers)` `= \ ^5P_3`
  `= 60`

 
ii.   `text(Number of different towers)`

`= \ ^5P_2 + \ ^5P_3 + \ ^5P_4 + \ ^5P_5`

`= 20 + 60 + 120 + 120`

`= 320`

Combinatorics, EXT1 A1 2015 HSC 14c

Two players `A` and `B` play a series of games against each other to get a prize. In any game, either of the players is equally likely to win.

To begin with, the first player who wins a total of 5 games gets the prize.

  1. Explain why the probability of player `A` getting the prize in exactly 7 games is  `((6),(4))(1/2)^7`.   (1 mark)

  2. Write an expression for the probability of player `A` getting the prize in at most 7 games.   (1 mark)

  3. Suppose now that the prize is given to the first player to win a total of `(n + 1)` games, where `n` is a positive integer.

     

    By considering the probability that `A` gets the prize, prove that
     

  4. `((n),(n))2^n + ((n + 1),(n))2^(n − 1) + ((n + 2),(n))2^(n − 2) + … + ((2n),(n)) = 2^(2n)`.   (2 marks)

Show Answers Only
  1. `text{Proof (See Worked Solutions)}`
  2. `\ ^4C_4(1/2)^5 +\ ^5C_4(1/2)^6 +\ ^6C_4(1/2)^7`
  3. `text{Proof (See Worked Solutions.)}`
Show Worked Solution

i.   `text(To win exactly 7 games, player)\ A`

♦♦♦ Mean mark 19%.

`text(must win the 7th game.)`
 

`:.P(A\ text{wins in 7 games)}`

`=\ ^6C_4 · (1/2)^4(1/2)^2 xx 1/2`

`=\ ^6C_4(1/2)^7`

 

ii.  `Ptext{(wins in at most 7 games)}`

♦♦♦ Mean mark 23%.

`=Ptext{(wins in 5, 6 or 7 games)}`

`=\ ^4C_4(1/2)^4 xx 1/2 +\ ^5C_4(1/2)^4(1/2) xx 1/2 +\ ^6C_4(1/2)^7`

`=\ ^4C_4(1/2)^5 +\ ^5C_4(1/2)^6 +\ ^6C_4(1/2)^7`

♦♦♦ Mean mark part (iii) 9%.

 

iii. `text(Prove that)`

`\ ^nC_n · 2^n +\ ^(n + 1)C_n · 2^(n-1) + … +\ ^(2n)C_n = 2^(2n)`

`P(A\ text(wins in)\ (n + 1)\ text{games)}`

`=\ ^nC_n(1/2)^(n + 1) +\ ^(n + 1)C_n(1/2)^(n + 2) + … +\ ^(2n)C_n(1/2)^(2n + 1)`
 

`text{One player must have won after (2n + 1) games are played.}`

`text(S)text(ince each player has an equal chance,)`
 

`\ ^nC_n(1/2)^(n + 1) +\ ^(n + 1)C_n(1/2)^(n + 2) + … +\ ^(2n)C_n(1/2)^(2n + 1) = 1/2`
 

`text(Multiply both sides by)\ 2^(2n + 1):`
 

`\ ^nC_n2^(-(n + 1)) · 2^(2n + 1) +\ ^(n + 1)C_n · 2^(-(n + 2)) · 2^(2n + 1) + …`

`… +\ ^(2n)C_n · 2^(-(2n + 1)) · 2^(2n + 1) = 2^(-1) · 2^(2n + 1)`
 

`:. \ ^nC_n · 2^n +\ ^(n + 1)C_n · 2^(n-1) + … +\ ^(2n)C_n = 2^(2n)`

Combinatorics, EXT1 A1 2015 HSC 4 MC

A rowing team consists of 8 rowers and a coxswain.

The rowers are selected from 12 students in Year 10.

The coxswain is selected from 4 students in Year 9.

In how many ways could the team be selected?

  1. `\ ^(12)C_8 +\ ^4C_1`
  2. `\ ^(12)P_8 +\ ^4P_1`
  3. `\ ^(12)C_8 ×\ ^4C_1`
  4. `\ ^(12)P_8 ×\ ^4P_1`
Show Answers Only

`C`

Show Worked Solution
 `\ ^(12)C_8` `=\ text(Combinations of rowers)`
 `\ ^4C_1` `=\ text(Combinations of coxswains)`

 
`:.\ text(Number of ways to select team)`

`=\ ^12C_8 xx\ ^4C_1`
 

`=> C`

Combinatorics, EXT1 A1 2008 HSC 4b

Barbara and John and six other people go through a doorway one at a time.

  1. In how many ways can the eight people go through the doorway if John goes through the doorway after Barbara with no-one in between?  (1 mark)

  2. Find the number of ways in which the eight people can go through the doorway if John goes through the doorway after Barbara.   (1 mark)

Show Answers Only
  1. `5040`
  2. `20\ 160`
Show Worked Solution

i.  `text(Barbara and John can be treated as one person)`

`:.\ text(# Combinations)` `= 7!`
  `= 5040`

 

ii.  `text(Solution 1)`

`text(Total possible combinations)=8!`

`text(Barbara has an equal chance of being behind as)`

`text(being in front of John.)`
 

`:.\ text{# Combinations (John before Barbara)}`

`=(8!)/2`

`=20\ 160`

 

`text(Solution 2)`

`text(If)\ B\ text(goes first,)\ J\ text(has 7 options and the other)`

`text(6 people can go in any order.)`

`=> text(# Combinations) = 7 xx 6!`
 

`text(If)\ B\ text(goes second,)\ J\ text(has 6 options)`

`=> text(# Combinations) = 6 xx 6!`
 

`text(If)\ B\ text(goes third,)\ J\ text(has 5 options)`

`=> text(# Combinations) = 5 xx 6!`

`text(And so on…)`
 

`:.\ text(Total Combinations)`

`= 7 xx 6! + 6 xx 6! + … + 1 xx 6!`

`= 6! ( 7 + 6 + 5 + 4 + 3 + 2 + 1)`

`= 6! ( 28)`

`= 20\ 160`

Combinatorics, EXT1 A1 2014 HSC 14b

Two players  `A`  and  `B`  play a game that consists of taking turns until a winner is determined. Each turn consists of spinning the arrow on a spinner once. The spinner has three sectors  `P`,  `Q`  and  `R`. The probabilities that the arrow stops in sectors  `P`,  `Q` and  `R`  are  `p`,  `q`  and  `r`  respectively.
 

2014 14b
 

The rules of the game are as follows:

• If the arrow stops in sector  `P`, then the player having the turn wins.

• If the arrow stops in sector  `Q`, then the player having the turn loses and the other player wins.

• If the arrow stops in sector  `R`, then the other player takes a turn.

Player  `A`  takes the first turn.

  1. Show that the probability of player  `A`  winning on the first or the second turn of the game is  `(1 − r) (p + r)`.   (2 marks)

  2. Show that the probability that player  `A`  eventually wins the game is  `(p + r)/(1 + r)`.   (3 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution

i.  `text(Show)\ P text{(}A\ text(wins)\ T_1\ text(or)\ T_2 text{)} = (1 – r)(p + r)`

`P (A\ text(wins on)\ T_1) = p`

`P (A\ text(wins on)\ T_2)`

`= P text{(} A\ text(lands on)\ R,\ B\ text(lands on)\ Q text{)}`

`= rq`

 

`:.\ P text{(}A\ text(wins)\ T_1\ text(or)\ T_2 text{)}`

♦♦ Mean mark 25%.
HINT: Expand out the solution `(1-r)(p+r)“=p+r-rp-r^2` to get a better idea of what you need to prove. 

`= p + rq`
 

`text(S)text(ince)\ q = 1 – (p + r),`

`P text{(} A\ text(wins)\ T_1\ text(or)\ T_2 text{)}`

`= p + r [1 – (p + r)]`

`= p + r – rp – r^2`

`= (1 – r)(p + r)\ \ \ text(… as required.)`
 

ii.  `text(Show)\ P text{(}A\ text(wins eventually) text{)} = (p + r)/(1 + r)`

 

`P text{(} A\ text(wins)\ T_1\ text(or)\ T_2 text{)} = (1 – r)(p + r)`

`P text{(} text(No result)\ T_1\ text(or)\ T_2 text{)} = r * r = r^2`
 

`:.\ P text{(} A\ text(wins eventually) text{)}`

`= underbrace{(1 – r)(p + r) + r^2 (1 – r)(p + r) + r^2*r^2 (1 – r)(p + r) + …}_{text(GP where)\ \ a = (1 – r)(p + r),\ \ r = r^2}`
 

`text(S)text(ince)\ \ 0 < r < 1 \ \ =>\ \  0 < r^2 < 1`

♦♦♦ Mean mark 11%. Lowest in the 2014 HSC exam!
HINT: The use of ‘eventually’ in the question should flag the possibility of solving by using `S_oo` in a GP.
`:.\ S_oo` `= a/(1 – r)`
  `= ((1 – r)(p + r))/(1 – r^2)`
  `= ((1 – r)(p + r))/((1 – r)(1 + r))`
  `= (p + r)/(1 + r)\ \ \ text(… as required)`

Combinatorics, EXT1 A1 2014 HSC 8 MC

In how many ways can 6 people from a group of 15 people be chosen and then arranged
in a circle?

  1. `(14!)/(8!)`
  2. `(14!)/(8! 6)`
  3. `(15!)/(9!)`
  4. `(15!)/(9! 6)`
Show Answers Only

`D`

Show Worked Solution

`text(# Arrangements)`

`=\ ^15C_6 xx 5!`

`= (15! 5!)/(6! 9!)`

`= (15!)/(9! 6)`

 
`=>  D`

Combinatorics, EXT1 A1 2013 HSC 7 MC

A family of eight is seated randomly around a circular table. 

What is the probability that the two youngest members of the family sit together?

  1. `(6!\ 2!)/(7!)`
  2. `(6!)/(7!\ 2!)`
  3. `(6!\ 2!)/(8!)`
  4. `(6!)/(8!\ 2!)` 
Show Answers Only

`A`

Show Worked Solution

`text(Fix youngest person in 1 seat,)`

`text(Total combinations around table) = 7!`

`text(Combinations with youngest side by side) =2!6!`
 

`:.\ text{P(sit together)} = (6!\ 2!)/(7!)`

`=>  A`

Combinatorics, EXT1 A1 2010 HSC 3a

At the front of a building there are five garage doors. Two of the doors are to be painted red, one is to be painted green, one blue and one orange. 

  1. How many possible arrangements are there for the colours on the doors?   (1 mark)

  2. How many possible arrangements are there for the colours on the doors if the two red doors are next to each other?    (1 mark)

Show Answers Only
  1. `60`
  2. `24`
Show Worked Solution
i.    `text(# Arrangements)` `= (5!)/(2!)`
    `= 60`

 

♦ Mean mark 50%
MARKER’S COMMENT: Drawing a diagram was a successful strategy for many students in this part.
ii.    `text(When 2 red doors are side-by-side,)`
`text(# Arrangements)` `= 4!`
  `= 24`

Combinatorics, EXT1 A1 2011 HSC 2e

Alex’s playlist consists of 40 different songs that can be arranged in any order.  

  1. How many arrangements are there for the 40 songs?    (1 mark)

  2. Alex decides that she wants to play her three favourite songs first, in any order.
  3. How many arrangements of the 40 songs are now possible?   (1 mark)

Show Answers Only
  1. `40!`
  2. `6 xx 37!`
Show Worked Solution
i.    `#\ text(Arrangements) = 40!`

 

ii.    `#\ text(Arrangements)` `= 3! xx 37!`
    `= 6 xx 37!`

Combinatorics, EXT1 A1 2012 HSC 11e

In how many ways can a committee of 3 men and 4 women be selected from a group of 8 men and 10 women?     (1 mark)

Show Answers Only

`11\ 760`

Show Worked Solution
`text(# Combinations)` `=\ ^8C_3 xx\ ^10C_4`
  `= (8!)/(5!3!) xx (10!)/(6!4!)`
  `= 56 xx 210`
  `= 11\ 760`

Combinatorics, EXT1 A1 2012 HSC 5 MC

How many arrangements of the letters of the word  `OLYMPIC`  are possible if the  `C`  and  the  `L`  are to be together in any order?

  1. `5!`  
  2. `6!` 
  3. `2 xx 5!` 
  4. `2 xx 6!` 
Show Answers Only

`D`

Show Worked Solution

`text(S)text(ince)\ C\ text(and)\ L\ text(must be kept together, they)`

`text(act as 1 letter with 2 possible combinations.)`
 

`:.\ text(Total combinations)`

`= 2 xx 6 xx 5 xx 4 xx 3 xx 2 xx 1`

`= 2 xx 6!`
 

`=>  D`