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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 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 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 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 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`