(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:

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