Past Exam Questions

(ii) First, arrange the 7 cup cakes, which can be done in \(7!\) ways. This creates 8 gaps (including the ends) where the brownies can be placed. Choose 4 out of these 8 gaps to place the brownies, which can be done in \(^8P_4\) ways. Therefore, the total number of arrangements is \(7! \times ^8P_4 = 8467200\).

(iii) Treat the 4 chocolate biscuits as a single unit. This gives us 7 units to arrange: 1 chocolate unit, 6 shortbread biscuits, and 2 gingerbread biscuits. The total arrangements of these units is \(\frac{9!}{6! \times 2!}\), accounting for the identical shortbread and gingerbread biscuits. Therefore, the total number of arrangements is 252.

(i) To find the number of different lists Hannah can make, we use permutations since the order matters. The number of ways to choose and arrange 5 singers from 15 is given by the permutation formula:

\(^{15}P_5 = \frac{15!}{(15-5)!} = \frac{15!}{10!}\)

Calculating this gives:

\(15 \times 14 \times 13 \times 12 \times 11 = 360360\)

Thus, there are 360360 different lists.

(ii) For the specific order of performers (male, female, male, female, male), we calculate the number of ways to choose each performer:

- First performer (male): 5 choices

- Second performer (female): 10 choices

- Third performer (male): 4 choices (since one male is already chosen)

- Fourth performer (female): 9 choices (since one female is already chosen)

- Fifth performer (male): 3 choices (since two males are already chosen)

The total number of lists is:

\(5 \times 10 \times 4 \times 9 \times 3 = 5400\)

Thus, there are 5400 such lists.

Since Mark is already seated in the front of taxi P, we have 7 friends left to arrange.

Jon and Sarah are sitting in the back of taxi P, so we treat them as a single unit initially. This leaves us with 5 units to arrange: Jon-Sarah, 1 more person in the back of P, and 4 people in taxi Q.

First, choose 1 more person to sit in the back of taxi P from the remaining 5 friends: \\(\binom{5}{1} = 5 \\\).

Now, arrange Jon and Sarah in 2 ways (Jon-Sarah or Sarah-Jon): \\(2! = 2 \\\).

Arrange the 4 people in taxi Q: \\(4! = 24 \\\).

The total number of arrangements is: \\(5 \times 2 \times 24 = 240 \\\).

Since there are 2 possible ways to arrange Jon and Sarah, multiply by 2: \\(240 \times 2 = 480 \\\).

(ii) To have a duck at each end, choose 2 ducks from the 3 available. This can be done in \(^3P_2\) ways. The remaining 8 positions can be filled with the 10 remaining ornaments (6 wooden animals and 4 sea-shells) in \(^10P_8\) ways. Therefore, the total number of arrangements is \(^3P_2 \times ^{10}P_8 = 10886400\).

(iii) Place a duck at each end, which can be done in \(^3P_2 = 6\) ways. The remaining 8 positions must alternate between wooden animals and sea-shells. Arrange the 4 sea-shells in \(^4P_4 = 24\) ways and the 4 wooden animals in \(^6P_4 = 360\) ways. Since the arrangement must alternate, swap the positions of sea-shells and wooden animals in 2 ways. Therefore, the total number of arrangements is \(6 \times 24 \times 360 \times 2 = 103680\).

(i) Treat the 6 boys as a single unit or 'block'. This block can be arranged in the line with the 4 girls in (5 + 1)! = 6! ext{ ways.} Within the block, the 6 boys can be arranged among themselves in 6! ways. Thus, the total number of arrangements is 6! \times 5! = 86400.

(ii) To ensure no two girls stand next to each other, arrange the 6 boys first, which can be done in 6! ways. This creates 7 possible spaces (before, between, and after the boys) to place the 4 girls. Choose 4 out of these 7 spaces to place the girls, which can be done in \binom{7}{4} ways. Arrange the 4 girls in these spaces in 4! ways. Thus, the total number of arrangements is 6! \times \binom{7}{4} \times 4! = 604800.

First, arrange the 5 women. There are 5! ways to arrange them.

Next, consider the positions for the man. The man cannot be at either end, so he can be placed in any of the 4 middle positions.

Thus, the total number of arrangements is given by:

\(5! \times 4 = 120 \times 4 = 480\)

Alternatively, calculate the total number of arrangements of 6 people, which is 6!. Then subtract the cases where the man is at either end:

\(6! - 2 \times 5! = 720 - 240 = 480\)

(ii) Consider the groups: mountain bicycles, racing bicycle, ordinary bicycles, and spaces. There are 4 groups to arrange, which can be done in \(4!\) ways. The mountain bicycles can be arranged among themselves in \(3!\) ways, and the ordinary bicycles can be arranged in \(2!\) ways. Therefore, the total number of arrangements is \(4! \times 3! \times 2 = 288\).

(iii) The ordinary bicycles are at each end, so they can be arranged in \(2!\) ways. The remaining bicycles (3 mountain and 1 racing) can be arranged in \(4!\) ways. The spaces and bicycles form 5 groups, which can be arranged in \(5!\) ways. Therefore, the total number of arrangements is \(2! \times 4! \times 5 = 240\).

(ii) Treat the 2 spaniels as one block and the 2 retrievers as another block. This gives us 5 blocks to arrange: 1 spaniel block, 1 retriever block, and 3 individual poodles. The number of ways to arrange these 5 blocks is given by:

\(5! = 120\)

Within the spaniel block, the 2 spaniels can be arranged in \(2!\) ways, and within the retriever block, the 2 retrievers can be arranged in \(2!\) ways. Therefore, the total number of arrangements is:

\(5! \times 2! \times 2! = 120 \times 2 \times 2 = 480\)

(iii) First, arrange the 4 non-poodles (2 spaniels and 2 retrievers) in a line. The number of ways to do this is:

\(4! = 24\)

There are 5 gaps created by these 4 dogs (before the first dog, between each pair of dogs, and after the last dog). We need to place the 3 poodles in these gaps such that no two poodles are adjacent. This is equivalent to choosing 3 gaps out of the 5 available, which can be done in:

\(\binom{5}{3} = 10\)

Therefore, the total number of arrangements is:

\(4! \times \binom{5}{3} = 24 \times 10 = 240\)

However, the mark scheme indicates a different approach, leading to a total of:

\(1440\)

(ii) Treat each group of trees (hibiscus, jacaranda, oleander) as a single unit or 'super tree'. This gives us 3 'super trees' to arrange. The number of ways to arrange these 3 'super trees' is given by:

\(3!\)

Within each 'super tree', the trees can be arranged internally. The number of ways to arrange the hibiscus trees is \(4!\), the jacaranda trees is \(6!\), and the oleander trees is \(2!\). Therefore, the total number of arrangements is:

\(3! \times 4! \times 6! \times 2! = 207360\)

(iii) First, arrange the 8 non-hibiscus trees (6 jacaranda and 2 oleander) in a line. The number of ways to do this is:

\(8!\)

This creates 9 gaps (including the ends) where the hibiscus trees can be placed. We need to choose 4 out of these 9 gaps to place the hibiscus trees, ensuring no two hibiscus trees are adjacent. The number of ways to choose 4 gaps from 9 is given by:

\(\binom{9}{4}\)

Finally, arrange the 4 hibiscus trees within the chosen gaps. The number of ways to do this is:

\(4!\)

Therefore, the total number of arrangements is:

\(8! \times \binom{9}{4} \times 4! = 121,927,680\)

(i) Consider each family as a single unit or 'block'. There are 4 families, so we arrange these 4 blocks in different orders:

\(4! = 24 ext{ ways}\)

Within each family, the members can be arranged among themselves. The number of arrangements for each family is:

2! ext{ for the Lin family,}

3! ext{ for the O’Connor family,}

5! ext{ for the Ahmed family,}

2! ext{ for the Burton family.}

Thus, the total number of arrangements is:

\(4! imes 2! imes 3! imes 5! imes 2! = 829440\)

(ii) First, arrange the 8 children. There are:

8! ext{ ways to arrange the children.}

To ensure no two adults are consecutive, place the 6 adults in the 9 gaps created by the 8 children (before the first child, between children, and after the last child). Choose 6 out of these 9 gaps:

9 imes 8 imes 7 imes 6 imes 5 imes 4 ext{ ways to choose and arrange the gaps.}

Thus, the total number of arrangements is:

\(8! imes 9 imes 8 imes 7 imes 6 imes 5 imes 4 = 2438553600\)

(b) Treat each group (swimmers, cyclists, runners) as a single unit. This gives us 3 units to arrange, which can be done in \(3!\) ways. Within each group, the swimmers can be arranged in \(2!\) ways, the cyclists in \(3!\) ways, and the runners in \(4!\) ways. Therefore, the total number of arrangements is \(3! \times 2! \times 3! \times 4! = 6 \times 2 \times 6 \times 24 = 1728\).

(c) First, arrange the 6 non-cyclists (2 swimmers and 4 runners) in \(6!\) ways. This creates 7 gaps (before, between, and after these 6 people) where the 3 cyclists can be placed. Choose 3 out of these 7 gaps to place the cyclists, which can be done in \(\binom{7}{3}\) ways. The cyclists themselves can be arranged in \(3!\) ways. Therefore, the total number of arrangements is \(6! \times \binom{7}{3} \times 3! = 720 \times 35 \times 6 = 151200\).

(i) The number of ways to arrange 8 people in 12 seats with no restrictions is given by the permutation formula:

\(^{12}P_8 = \frac{12!}{(12-8)!} = 19,958,400\).

(ii) To ensure Mary and Frances do not sit next to each other, first calculate the total arrangements if they sit together. Treat Mary and Frances as a single unit, so there are 7 units to arrange in 11 seats:

\(^{11}P_7 = \frac{11!}{(11-7)!} = 1,663,200\).

Since Mary and Frances can switch places, multiply by 2:

\(1,663,200 \times 2 = 3,326,400\).

Subtract this from the total unrestricted arrangements:

\(19,958,400 - 3,326,400 = 16,632,000\).

(iii) If all 8 people sit together, treat them as a single block. There are 5 possible positions for this block in 12 seats. The number of ways to arrange the 8 people within the block is:

\(8! = 40,320\).

Multiply by the 5 positions:

\(40,320 \times 5 = 201,600\).

(a) Consider the packets of chocolates as a single unit. This gives us 5 packets of toffees, 4 packets of fruit gums, and 1 unit of chocolates, totaling 10 units to arrange. The number of arrangements is given by:

\(\frac{10!}{5!4!} = 1260\)

(b)(i) Jessica chooses 4 packets out of 8, considering the order. This is a permutation problem:

\({}_8P_4 = 8 \times 7 \times 6 \times 5 = 1680\)

(b)(ii) If the packets of chocolate biscuits and custard creams are chosen, we need to choose 2 more packets from the remaining 6. The number of ways to choose 2 packets from 6 is:

\(\binom{6}{2} = 15\)

Considering the order of 4 packets, we multiply by \(4!\):

\(15 \times 4! = 360\)

(i) Consider each couple as a single unit or 'block'. There are 7 such blocks. These blocks can be arranged in a line in \(7!\) ways. Within each block, the two people can be arranged in \(2!\) ways. Therefore, the total number of arrangements is \(7! \times 2^7 = 645,120\).

(ii) Treat the 7 friends as one group and the 7 partners as another group. These two groups can be arranged in \(2!\) ways. Within each group, the individuals can be arranged in \(7!\) ways. Therefore, the total number of arrangements is \(7! \times 7! \times 2 = 50,803,200\).

To solve this problem, we treat each folder as a single unit or block. Thus, we have 3 blocks: Family, Holiday, and Friends.

The number of ways to arrange these 3 blocks is given by the factorial of the number of blocks:

\(3! = 6\)

Within each block, the images can be arranged among themselves. Therefore, we calculate the factorial for each folder:

Family folder: \(3! = 6\)

Holiday folder: \(4! = 24\)

Friends folder: \(8! = 40,320\)

The total number of arrangements is the product of these factorials:

\(3! \times 3! \times 4! \times 8! = 6 \times 6 \times 24 \times 40,320 = 34,836,480\)

1. Each coin can show either a head or a tail, giving 2 possibilities per coin. For 12 coins, the total number of arrangements is given by:

   \(2^{12} = 4096\)

2. To find the number of arrangements with 7 heads and 5 tails, use the combination formula:

   \(\binom{12}{7} = \frac{12!}{7!5!} = 792\)

(i) When the order of choosing is taken into account, we need to multiply the number of combinations by the number of permutations of the 4 astronauts. The number of permutations of 4 astronauts is given by:

\(4! = 4 \times 3 \times 2 \times 1 = 24\)

Thus, the total number of ways is:

\(3876 \times 24 = 93024\)

(ii) Each astronaut has 3 possessions, and these possessions are kept together. We treat each astronaut's possessions as a single unit, so we have 4 units to arrange. The number of ways to arrange these 4 units is:

\(4! = 24\)

Within each unit, the 3 possessions can be arranged in:

\(3! = 6\) ways

Therefore, the total number of ways to arrange the possessions is:

\((3!)^4 \times 4! = 6^4 \times 24 = 31104\)

(i) To find the number of different outfits, multiply the number of choices for each item: shoes, jeans, and tee shirts. Thus, the total number of outfits is:

\(4 \times 3 \times 7 = 84\)

(ii) First, calculate the total number of arrangements of 10 items (3 jeans + 7 tee shirts):

\(10! = 3628800\)

Next, calculate the arrangements where the two blue items are together. Treat the two blue items as a single unit, reducing the problem to arranging 9 items:

\(9! = 362880\)

Within this unit, the two blue items can be arranged in \(2!\) ways:

\(2! = 2\)

Thus, the number of arrangements where the blue items are together is:

\(9! \times 2 = 725760\)

Subtract this from the total arrangements to find the number of ways the blue items are not together:

\(10! - 9! \times 2 = 2903040\)

(iii) Fahad can choose any number of books from 1 to 9. The number of ways to choose at least one book is the total number of subsets of 9 books minus the empty set:

The total number of subsets is \(2^9\):

\(2^9 = 512\)

Subtract the empty set:

\(512 - 1 = 511\)

(i) The word HAPPINESS consists of 9 letters where P appears twice and S appears twice. The number of different arrangements is given by:

\(\frac{9!}{2! \times 2!} = \frac{362880}{4} = 90720\)

(ii) To find the probability that the 3 vowels (A, E, I) are not all next to each other, first calculate the number of arrangements where they are together. Treat the 3 vowels as a single unit, so we have 7 units to arrange: (AEI), H, P, P, N, S, S.

The number of arrangements of these 7 units is:

\(\frac{7!}{2! \times 2!} = \frac{5040}{4} = 1260\)

Within the unit (AEI), the vowels can be arranged in \(3!\) ways:

\(3! = 6\)

Thus, the total number of arrangements where the vowels are together is:

\(1260 \times 6 = 7560\)

The probability that the vowels are not all together is:

\(1 - \frac{7560}{90720} = \frac{83160}{90720} = 0.917 \left(=\frac{11}{12}\right)\)

(i) To find the number of possible seating arrangements for the 12 passengers out of 14 seats, we use permutations:

\(^{14}P_{12} = \frac{14!}{(14-12)!} = \frac{14!}{2!} = 4.36 \times 10^{10}\).

(ii) For the specific seating arrangement:

1. The 3 business people can be arranged in the front row in \(3! = 6\) ways.
2. The 5 students each sit at a window seat, which can be arranged in \(5! = 120\) ways.
3. Mr and Mrs Lin can sit in the same row on the same side of the aisle in \(3P2 \times 2 = 12\) ways (choosing 2 out of 3 rows and 2 ways to arrange them).
4. Mr and Mrs Brown can sit in another row on the same side of the aisle in \(2 \times 2 = 4\) ways.
5. Total arrangements = \(6 \times 120 \times 12 \times 4 = 17280\).