How many QCAA Specialist Mathematics Alternative Sequence questions cover Combinatorics?

AusGrader has 8 QCAA Specialist Mathematics Alternative Sequence questions on Combinatorics, all with instant AI grading and detailed marking feedback.

QCAA Specialist Mathematics Alternative Sequence — Combinatorics Questions & Answers

Q4

2021

QCAA

Paper 2

1 mark

Q4

1 mark

New Queensland vehicle numberplates have three digits followed by two letters and then another digit, e.g. 999 AZ9.

If digits and letters can be repeated, how many different numberplates are possible?

A

17 576 000

B

6 760 000

C

4 435 236

D

3 276 000

Reveal Answer

A

17 576 000

This incorrect option calculates the number of combinations for three letters and three digits ( $26^3 \times 10^3 = 17,576,000$ ), rather than the required two letters and four digits.

B

6 760 000

Correct Answer

There are 10 possible choices (0-9) for each of the four digits and 26 possible choices (A-Z) for each of the two letters. Since repetition is allowed, the total number of combinations is $10^4 \times 26^2 = 6,760,000$ .

C

4 435 236

This incorrect option calculates the combinations assuming there are only 9 possible choices for the digits (e.g., excluding zero), which would result in $9^4 \times 26^2 = 4,435,236$ .

D

3 276 000

This incorrect option calculates the number of combinations if digits and letters CANNOT be repeated ( $10 \times 9 \times 8 \times 26 \times 25 \times 7 = 3,276,000$ ).

Q4

2021

QCAA

Paper 1

1 mark

Q4

1 mark

The value of ${}^6C_2-{}^3P_2$ is

A

6

B

9

C

12

D

15

Reveal Answer

A

6

This is incorrect because 6 is the value of ${}^3P_2$ alone, rather than the difference between the two terms.

B

9

Correct Answer

This is correct. Using the formulas for combinations and permutations, ${}^6C_2 = 15$ and ${}^3P_2 = 6$ . Subtracting them gives $15 - 6 = 9$ .

C

12

This is incorrect and likely the result of an arithmetic error. The correct values are ${}^6C_2 = 15$ and ${}^3P_2 = 6$ , which subtract to 9.

D

15

This is incorrect because 15 is the value of ${}^6C_2$ alone, rather than the difference between the two terms.

Q1

2021

QCAA

Paper 2

1 mark

Q1

1 mark

Two hundred people were surveyed about whether they owned a cat or a dog. The table shows the results of the survey.

Pet owned	Cat	Dog	Neither
Number of people	105	108	7

The number of people who owned both a cat and a dog is

A

6

B

7

C

13

D

20

Reveal Answer

A

6

This value does not follow from the data provided. The correct approach requires using the Principle of Inclusion-Exclusion.

B

7

This is the number of people who own neither a cat nor a dog, not the number of people who own both.

C

13

This calculation ( $105 + 108 - 200 = 13$ ) incorrectly assumes everyone owns at least one pet, forgetting to account for the 7 people who own neither.

D

20

Correct Answer

Since 7 people own neither, $200 - 7 = 193$ people own at least one pet. Using the Principle of Inclusion-Exclusion, the number of people who own both is $(105 + 108) - 193 = 20$ .

Q19

2021

QCAA

Paper 2

7 marks

Q19

7 marks

A game uses four dice with six faces numbered 0, 1, 2, 3, 4 and 5. Each number is equally likely to occur. Three dice are the same size, while the fourth is larger.

The game involves a player rolling the four dice simultaneously. A player wins the game if the sum of the numbers on the three smaller dice equals the number on the larger die. Determine the probability of a player winning at least twice in 10 games.

Reveal Answer

Winning results

Large die value	Small dice values
0	0, 0, 0
1	0, 0, 1
2	0, 0, 2 or 0, 1, 1
3	0, 0, 3 or 0, 1, 2 or 1, 1, 1
4	0, 0, 4 or 0, 1, 3 or 0, 2, 2 or 1, 1, 2
5	0, 0, 5 or 0, 1, 4 or 0, 2, 3 or 1, 1, 3 or 1, 2, 2

Number of ways each result can occur

Large die value	Small dice values	Number of ways each result can occur
0	0, 0, 0	$\frac{3!}{3!} = 1$
1	0, 0, 1	$\frac{3!}{2!} = 3$
2	0, 0, 2 0, 1, 1	$\frac{3!}{2!} = 3$ $\frac{3!}{2!} = 3$
3	0, 0, 3 0, 1, 2 1, 1, 1	$\frac{3!}{2!} = 3$ $3! = 6$ $\frac{3!}{3!} = 1$
4	0, 0, 4 0, 1, 3 0, 2, 2 1, 1, 2	$\frac{3!}{2!} = 3$ $3! = 6$ $\frac{3!}{2!} = 3$ $\frac{3!}{2!} = 3$
5	0, 0, 5 0, 1, 4 0, 2, 3 1, 1, 3 1, 2, 2	$\frac{3!}{2!} = 3$ $3! = 6$ $3! = 6$ $\frac{3!}{2!} = 3$ $\frac{3!}{2!} = 3$

P(winning a game)
$= \frac{1 + 3 + 6 + 10 + 15 + 21}{6^4} = \frac{7}{162}$

Using binomial distribution on GDC
P(winning at least twice in 10 games) $\approx 0.07$

Marking Criteria

Descriptor	Marks
correctly identifies all possible smaller dice value combinations that produce the large die values of 0, 1 and 2	1
correctly identifies all possible smaller dice value combinations that produce the large die values of 3, 4 and 5	1
determines number of arrangements that produce the large die values of 0, 1 and 2	1
determines number of arrangements that produce the large die values of 3, 4 and 5	1
determines probability of winning a game	1
determines required probability	1
shows logical organisation, communicating key steps to at least where the number of arrangements that produce all possible large die values are determined	1

Q6

2021

QCAA

Paper 1

1 mark

Q6

1 mark

A game is played on an $8\times 8$ square board. Each square on the board can be identified by a letter and a number. The square A1 is shaded.

A player starts at A1 and can move either one square up or one square right at a time.

The number of ways that each square on the board can be reached from A1 has been partially completed on the diagram.

	A	B	C	D	E	F	G	H
8	1
7	1
6	1
5	1	5			?
4	1	4	10
3	1	3	6	10
2	1	2	3	4	5
1	■	1	1	1	1	1	1	1

Determine the number of ways a player can reach E5.

A

$\binom{7}{4}$

B

$\binom{7}{5}$

C

$\binom{8}{4}$

D

$\binom{8}{5}$

Reveal Answer

A

$\binom{7}{4}$

This is incorrect. $\binom{7}{4}$ represents the number of paths requiring exactly 7 total moves, such as reaching D5 (3 right, 4 up) or E4 (4 right, 3 up).

B

$\binom{7}{5}$

This is incorrect. $\binom{7}{5}$ represents paths with 7 total moves where 5 are in one direction, which does not match the 8 total moves needed to reach E5.

C

$\binom{8}{4}$

Correct Answer

This is correct. To reach E5 from A1, the player must move exactly 4 squares right (A to E) and 4 squares up (1 to 5). The number of unique paths is the number of ways to choose 4 right moves out of 8 total moves, which is $\binom{8}{4}$ .

D

$\binom{8}{5}$

This is incorrect. $\binom{8}{5}$ represents paths with 8 total moves where 5 are in one direction, such as reaching F4 (5 right, 3 up) or D6 (3 right, 5 up).

Q12

2021

QCAA

Paper 2

6 marks

Q12a

1 mark

Determine the number of different ways all the letters of the word BEEKEEPER can be arranged if all the Es are together.

Reveal Answer

Number of ways $= 120$

Marking Criteria

Descriptor	Marks
correctly determines the number of arrangements	1

Q12b

2 marks

Determine the number of different ways all the letters of the word BEEKEEPER can be arranged if E is at one end and P is at the other end.

Reveal Answer

Number of ways with E first and P last $= \frac{7!}{4!}$

Total number of ways $= 2 \times \frac{7!}{4!} = 420$

Marking Criteria

Descriptor	Marks
correctly determines that there are $\frac{7!}{4!}$ ways the letters can be arranged with E first and P last (or vice versa)	1
recognises that a factor of 2 is required to determine the total number of ways	1

Q12c

3 marks

If four letters are randomly selected from the word BEEKEEPER, determine the number of selections that contain one or two Es.

Reveal Answer

Number of selections with 1 E $= ^5C_1 \ ^4C_3 = 20$

Number of selections with 2 Es $= ^5C_2 \ ^4C_2 = 60$

Total number of selections with either 1 or 2 Es
$= 20 + 60 = 80$

Marking Criteria

Descriptor	Marks
correctly uses the multiplication principle to determine the number of selections with 1 E	1
correctly uses the multiplication principle to determine the number of selections with 2 Es	1
uses addition principle to determine the total number of selections with 1 E or 2 Es	1

QCAA Specialist Mathematics Alternative Sequence Combinatorics

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