Past Exam Questions

(i) To find the number of different arrangements of 6 pegs, all of different colours, in 4 holes, we use permutations. The number of ways to choose 4 pegs from 6 is given by the permutation formula:

\(^6P_4 = \frac{6!}{(6-4)!} = \frac{6!}{2!} = 360\)

Thus, there are 360 different arrangements.

(ii) For 4 pegs consisting of 2 blue pegs, 1 orange peg, and 1 yellow peg, we need to account for the repetition of the blue pegs. The formula for permutations of multiset is:

\(\frac{4!}{2!} = 12\)

Therefore, there are 12 different arrangements.

Consider Abel and Betty as a single unit or block. This reduces the problem to arranging 6 units: (AB), Cally, Doug, Eve, Freya, and Gino.

The number of ways to arrange these 6 units is given by:

\(6! = 720\)

Within the block (AB), Abel and Betty can be arranged in:

\(2! = 2\)

Thus, the total number of arrangements with Abel and Betty together is:

\(6! \times 2! = 720 \times 2 = 1440\)

Now, consider the arrangements where Freya and Gino are also together. Treat Freya and Gino as another block, reducing the problem to arranging 5 units: (AB), (FG), Cally, Doug, and Eve.

The number of ways to arrange these 5 units is:

\(5! = 120\)

Within the block (FG), Freya and Gino can be arranged in:

\(2! = 2\)

Thus, the total number of arrangements with both (AB) and (FG) together is:

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

Therefore, the number of arrangements where Abel and Betty are together but Freya and Gino are not together is:

\(1440 - 480 = 960\)

(i) Consider the cases where the first and last cans are the same type:

• If they are cola, the arrangement is \\frac{5!}{2!2!} = 30\ ways.
• If they are green tea, the arrangement is \\frac{5!}{3!2!} = 10\ ways.
• If they are orange juice, the arrangement is \\frac{5!}{3!2!} = 10\ ways.

\(Total arrangements = 30 + 10 + 10 = 50\ ways.\)

\((ii) Treat the 3 cans of cola as a single block. The arrangement without restriction is \\frac{5!}{2!2!} = 30\ ways.\)

\(Now consider the case where both colas and green teas are together: \\frac{4!}{2!} = 12\ ways.\)

\(Subtract the restricted case from the total: 30 - 12 = 18\ ways.\)

(i) The total number of arrangements of 9 cards is given by the factorial of 9, which is:

\(9! = 362880\)

(ii) To find the number of arrangements where the pink and green cards are not next to each other, first calculate the total arrangements where they are together. Treat the pink and green cards as a single unit, so there are 8 units to arrange:

\(8! = 40320\)

Within this unit, the pink and green cards can be arranged in 2 ways (PG or GP):

\(8! \times 2 = 80640\)

Subtract this from the total arrangements:

\(362880 - 80640 = 282240\)

(iii) The number of ways to choose and arrange 3 cards from 9 is given by:

\(^9P_3 = \frac{9!}{(9-3)!} = 504\)

(iv) To find the number of arrangements containing the pink card, choose 2 more cards from the remaining 8 and arrange them:

\(^8C_2 \times 3! = 28 \times 6 = 168\)

(v) For arrangements not having the pink card next to the green card, calculate the arrangements where they are together and subtract from the total in part (iii). Treat PG as a single unit, choose 1 more card from the remaining 7, and arrange:

\(7 \times 2 \times 2 = 28\)

Subtract from total arrangements of 3 cards:

\(504 - 28 = 476\)

(ii) To arrange the 9 mugs such that the 3 china mugs are separated, first arrange the 6 plastic mugs. There are \(6!\) ways to arrange these. This creates 7 gaps (before, between, and after the plastic mugs) to place the china mugs. Choose 3 out of these 7 gaps, which can be done in \(\binom{7}{3}\) ways. Arrange the 3 china mugs in these gaps in \(3!\) ways. Therefore, the total number of arrangements is \(6! \times \binom{7}{3} \times 3! = 720 \times 35 \times 6 = 151200\).

(iii) Treat the 3 identical red mugs as a single unit. This gives us 12 units to arrange: 1 red unit, 4 blue mugs, and 7 yellow mugs. The total number of arrangements is \(\frac{12!}{4! \times 7!}\), accounting for the identical blue and yellow mugs. Therefore, the number of arrangements is \(\frac{479001600}{24 \times 5040} = 3960\).

(ii) Consider the 6 tallest sopranos as a single block. This block can be arranged in a line with the remaining 4 sopranos. There are 5 blocks in total (1 block of 6 sopranos and 4 individual sopranos). The number of ways to arrange these 5 blocks is given by:

\(5!\)

Within the block of 6 sopranos, they can be arranged among themselves in:

\(6!\)

Therefore, the total number of arrangements is:

\(5! \times 6! = 86400\)

(iii) Consider the 4 tenors as a single block and the 4 basses as another block. These two blocks can be arranged in:

\(2!\)

Within the block of 4 tenors, they can be arranged among themselves in:

\(4!\)

Within the block of 4 basses, they can be arranged among themselves in:

\(3!\)

Therefore, the total number of arrangements is:

\(4! \times 3! \times 2 = 288\)

(i) To find the number of possible arrangements of the 12 houses, we use the formula for permutations of multiset:

\(\frac{12!}{2!2!3!4!}\)

Calculating this gives:

\(\frac{479001600}{4 \times 2 \times 6 \times 24} = 831600\)

(ii) For the second part, we need to arrange the houses in two groups of 6. In the first group, we have all houses in styles A and D, which are 2 and 4 respectively. The number of arrangements for the first group is:

\(\frac{6!}{4!2!}\)

In the second group, we have all houses in styles B, C, and E, which are 2, 3, and 1 respectively. The number of arrangements for the second group is:

\(\frac{6!}{2!3!}\)

The total number of arrangements is the product of the arrangements of the two groups:

\(\frac{6!}{4!2!} \times \frac{6!}{2!3!} = 900\)

To solve this problem, treat the 3 jazz CDs as a single unit or 'block'. This reduces the problem to arranging 9 units: 6 pop CDs, 2 classical CDs, and 1 jazz block.

The number of ways to arrange these 9 units is given by:

\(9!\)

Within the jazz block, the 3 jazz CDs can be arranged among themselves in:

\(3!\)

Therefore, the total number of arrangements is:

\(9! \times 3!\)

Calculating these factorials:

\(9! = 362,880\)

\(3! = 6\)

Thus, the total number of arrangements is:

\(362,880 \times 6 = 2,177,280\)

Alternatively, the mark scheme suggests another possible answer of 2,180,000, which might be due to rounding or approximation in a different context.

(i) There are 9 people in total. The number of possible arrangements with no restrictions is given by the factorial of the total number of people:

\(9! = 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 362880\).

(ii) First, arrange the 6 men. The number of ways to do this is \(6!\):

\(6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720\).

There are 7 gaps created between and at the ends of the men where the women can be placed (before the first man, between men, and after the last man). We need to choose 3 of these 7 gaps to place the women, ensuring no two women are adjacent. The number of ways to choose 3 gaps from 7 is given by \(\binom{7}{3}\), and the number of ways to arrange the 3 women in these gaps is \(3!\):

\(\binom{7}{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35\).

\(3! = 3 \times 2 \times 1 = 6\).

The total number of arrangements is \(6! \times \binom{7}{3} \times 3! = 720 \times 35 \times 6 = 151200\).

(i) To find the number of possible seating arrangements for the 11 passengers, we use the permutation formula \(^nP_r\), where \(n\) is the total number of seats and \(r\) is the number of passengers. Here, \(n = 17\) and \(r = 11\).

\(^{17}P_{11} = \frac{17!}{(17-11)!} = \frac{17!}{6!}\)

Calculating this gives \(4.94 \times 10^{11}\).

(ii) If 5 particular people sit in the back row, we first arrange these 5 people in the 5 seats of the back row, which can be done in \(5!\) ways.

Then, we arrange the remaining 6 passengers in the remaining 12 seats. This can be done in \(^{12}P_6\) ways.

Thus, the total number of arrangements is \(5! \times ^{12}P_6\).

\(5! = 120\)

\(^{12}P_6 = \frac{12!}{(12-6)!} = \frac{12!}{6!}\)

Calculating this gives \(79800000\).

(i) To find the number of different arrangements for parking the 9 cars and leaving 4 empty spaces, we use permutations. The total number of ways to arrange 9 cars in 13 spaces is given by the permutation formula:

\(^{13}P_9 = \frac{13!}{(13-9)!} = \frac{13!}{4!} = 259,459,200\).

(ii) If the 4 empty spaces are next to each other, treat them as a single block. This reduces the problem to arranging 10 blocks (9 cars + 1 block of empty spaces). The number of arrangements is:

\(10! = 3,628,800\).

(iii) The probability that there will not be 4 empty spaces next to each other is the complement of the probability that they are next to each other. Thus,

\(P(\text{not next to each other}) = 1 - \frac{\text{arrangements with 4 empty spaces together}}{\text{total arrangements}}\).

\(P = 1 - \frac{3,628,800}{259,459,200} = 0.986\).

Since Mr Lizo goes first, we consider the remaining families as groups:

• Lizo family: 4 members (excluding Mr Lizo)
• Kenny family: 6 members
• Martin family: 2 members
• Nantes family: 2 members

First, calculate the number of ways to arrange the families:

\(3! = 6 ext{ ways (for Kenny, Martin, Nantes)}\)

Next, calculate the arrangements within each family:

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

\(6! = 720 ext{ ways (Kenny family)}\)

\(2! = 2 ext{ ways (Martin family)}\)

\(2! = 2 ext{ ways (Nantes family)}\)

Multiply these together to find the total number of arrangements:

\(4! imes 6! imes 2! imes 2! imes 3! = 24 imes 720 imes 2 imes 2 imes 6 = 414,720\)

Consider the 3 boys as a single unit or 'block'. This reduces the problem to arranging 10 units: 1 block of boys, 4 girls, and 5 adults.

Since an adult must be at each end, we choose 2 adults to be at the ends. The number of ways to choose 2 adults from 5 is given by:

\(^5P_2 = 5 \times 4 = 20\)

Now, arrange the remaining 8 units (1 block of boys, 4 girls, and 3 adults) in the middle:

\(8! = 40,320\)

Within the block of boys, the 3 boys can be arranged among themselves in:

\(3! = 6\)

Therefore, the total number of arrangements is:

\(8! \times 3! \times ^5P_2 = 40,320 \times 6 \times 20 = 4,838,400\)

(a) The number of ways to choose 2 letters from 5 is given by permutations: \(^5P_2 = 5 \times 4\). The number of ways to choose 4 digits from 7 is \(^7P_4 = 7 \times 6 \times 5 \times 4\). Therefore, the total number of codes is \(5 \times 4 \times 7 \times 6 \times 5 \times 4 = 16,800\).

(b) We use the principle of inclusion-exclusion. Calculate the number of codes with A, with 5, and with both:

1. Codes with A and no 5: \(8 \times ^6P_4 = 2,880\).
2. Codes with 5 and no A: \(4 \times ^6P_3 = 5,760\).
3. Codes with both A and 5: \(8 \times ^5P_3 = 3,840\).

Total codes with A or 5 or both: \(2,880 + 5,760 + 3,840 = 12,480\).

(c) The number of codes that start with DE and have a number between 4500 and 5000 is calculated as follows:

There is only one way to choose DE. The number must be between 4500 and 5000, so the first digit is 4, and the remaining digits are chosen from \{5, 6, 7\}:

Number of ways to choose the remaining 3 digits: \(^3P_3 = 6\).

Total successful codes: \(1 \times 6 = 6\).

Probability: \(\frac{6}{16,800} = \frac{1}{280}\).

(a) To find the number of arrangements with Raman at the center, fix Raman in the center position. There are 8 remaining members to arrange around him. The number of ways to arrange these 8 members is given by the factorial of 8, which is:

\(8! = 40,320\)

(b) To find the number of arrangements where Raman and Sanjay are not next to each other, first calculate the total number of arrangements of the 9 members, which is \(9!\). Then, calculate the number of arrangements where Raman and Sanjay are together. Treat Raman and Sanjay as a single unit, so there are 8 units to arrange. The number of ways to arrange these 8 units is \(8!\), and within the unit, Raman and Sanjay can be arranged in \(2!\) ways. Thus, the number of arrangements where they are together is \(8! \times 2!\).

The number of arrangements where they are not together is:

\(9! - 8! \times 2! = 362,880 - 80,640 = 282,240\)

(c) The total number of arrangements of 9 people is given by the factorial of 9, which is:

\(9! = 362,880\)

To find the number of arrangements where Mr Ahmed and Mr Baker are not standing next to each other, we first calculate the arrangements where they are together. Treat Mr Ahmed and Mr Baker as a single unit, so we have 8 units to arrange. The number of arrangements for these 8 units is:

\(8! = 40,320\)

Within their unit, Mr Ahmed and Mr Baker can be arranged in 2 ways (Ahmed-Baker or Baker-Ahmed), so we multiply by 2:

\(8! \times 2 = 80,640\)

The number of arrangements where Mr Ahmed and Mr Baker are not together is:

\(9! - 8! \times 2 = 362,880 - 80,640 = 282,240\)

(d) To find the number of arrangements where there is exactly one person between Mr Ahmed and Mr Baker, consider the three people (Ahmed, Baker, and the person between them) as a single unit. There are 7 other people to arrange, so we have 7 units to arrange:

\(7! = 5,040\)

Within the unit, the arrangement can be Ahmed-Person-Baker or Baker-Person-Ahmed, so there are 2 ways to arrange them:

\(2\)

There are 7 choices for the person between Mr Ahmed and Mr Baker:

\(7\)

The total number of arrangements is:

\(7! \times 2 \times 7 = 5,040 \times 2 \times 7 = 70,560\)

There are 12 people in total, and they are divided into groups of 5, 4, and 3. We need to find the probability that Jai and Kaz are in the same group.

Method 1: Probabilities

Calculate the probability for each group:

• In the group of 5: \(\frac{5}{12} \times \frac{4}{11} = \frac{20}{132} = \frac{5}{33}\)
• In the group of 4: \(\frac{4}{12} \times \frac{3}{11} = \frac{12}{132} = \frac{1}{11}\)
• In the group of 3: \(\frac{3}{12} \times \frac{2}{11} = \frac{6}{132} = \frac{1}{22}\)

Add the probabilities for the three scenarios:

\(\frac{5}{33} + \frac{1}{11} + \frac{1}{22} = \frac{19}{66}\)

Method 2: Arrangements

Calculate the number of ways Jai and Kaz can be in each group:

• In the group of 5: \(\binom{10}{3} \times \binom{7}{4} = 4200\)
• In the group of 4: \(\binom{10}{2} \times \binom{8}{5} = 2520\)
• In the group of 3: \(\binom{10}{1} \times \binom{9}{5} = 1260\)

Total number of ways to arrange the groups: \(\binom{12}{5} \times \binom{7}{4} = 27720\)

Probability: \(\frac{4200 + 2520 + 1260}{27720} = \frac{7980}{27720} = \frac{19}{66}\)

(a) To choose a group of 6 with exactly 1 man, we select 1 man from 4 men and 5 women from 11 women. The number of ways to do this is given by:

\(\binom{4}{1} \times \binom{11}{5} = 4 \times 462 = 1848\)

(b) We need to find the number of ways to choose a group of 6 such that Jane and Kate are not both included. We can use two methods:

Method 1: Consider the scenarios:

1. Neither Jane nor Kate is selected: Choose 6 from the remaining 13 people.
2. Only Jane is selected: Choose 5 from the remaining 13 people.
3. Only Kate is selected: Choose 5 from the remaining 13 people.

The total number of ways is:

\(\binom{13}{6} + \binom{13}{5} + \binom{13}{5} = 1716 + 1287 + 1287 = 4290\)

Method 2: Calculate the total number of ways to choose 6 people from 15, then subtract the number of ways where both Jane and Kate are included:

\(\binom{15}{6} - \binom{13}{4} = 5005 - 715 = 4290\)

(c) To divide 9 members into a group of 5 and a group of 4, we choose 5 members out of 9. This can be calculated using combinations:

\(\binom{9}{5} = 126\)

(d) To find the probability that Raman and Sanjay are in the same group, consider two cases:

1. Both are in the group of 5. Choose 3 more members from the remaining 7:

\(\binom{7}{3} = 35\)

2. Both are in the group of 4. Choose 2 more members from the remaining 7:

\(\binom{7}{2} = 21\)

Total ways for both to be in the same group:

\(35 + 21 = 56\)

Probability that Raman and Sanjay are in the same group:

\(\frac{56}{126} = \frac{4}{9} \approx 0.444\)

To find the number of different selections of 5 marbles containing at least 4 marbles of the same color, we consider two cases:

1. 4 red marbles and 1 blue marble:

   The number of ways to choose 4 red marbles from 8 is given by \(\binom{8}{4}\), and the number of ways to choose 1 blue marble from 4 is \(\binom{4}{1}\).

   Thus, the total number of ways for this case is \(\binom{8}{4} \times \binom{4}{1} = 70 \times 4 = 280\).

2. 4 blue marbles and 1 red marble:

   The number of ways to choose 4 blue marbles from 4 is \(\binom{4}{4}\), and the number of ways to choose 1 red marble from 8 is \(\binom{8}{1}\).

   Thus, the total number of ways for this case is \(\binom{4}{4} \times \binom{8}{1} = 1 \times 8 = 8\).

3. 5 red marbles:

   The number of ways to choose 5 red marbles from 8 is \(\binom{8}{5}\).

   Thus, the total number of ways for this case is \(\binom{8}{5} = 56\).

Adding all these cases together, the total number of selections is:

\(280 + 8 + 56 = 344\).