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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 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 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 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 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 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!`