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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 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 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 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 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 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`