(a) The word TELESCOPE consists of 9 letters where E appears twice. The number of different arrangements is given by:

\(\frac{9!}{2!} = \frac{362,880}{2} = 181,440\)

However, the mark-scheme indicates the correct answer is 60,480, which suggests a different interpretation or correction in the problem setup.

(b) To arrange the letters such that there are exactly two letters between T and C, consider T and C as a block with two spaces between them. This block can be arranged in 7 positions among the 9 letters:

\(\frac{7!}{3!} \times 2 \times 6 = 10,080\)

The factor of 2 accounts for the two possible orders of T and C within the block.

The word REQUIREMENT consists of the letters: R, E, Q, U, I, R, E, M, E, N, T.

We need to select 5 letters containing at least two Es and at least one R.

Consider the following cases:

1. Case 1: 2 Es and 1 R

We have E, E, R, and need to choose 2 more letters from the remaining 8 letters (Q, U, I, M, N, T, R, E).

The number of ways to choose 2 more letters is given by: \(\binom{8}{2} = 28\).

2. Case 2: 3 Es and 1 R

We have E, E, E, R, and need to choose 1 more letter from the remaining 7 letters (Q, U, I, M, N, T, R).

The number of ways to choose 1 more letter is given by: \(\binom{7}{1} = 7\).

3. Case 3: 2 Es and 2 Rs

We have E, E, R, R, and need to choose 1 more letter from the remaining 7 letters (Q, U, I, M, N, T, E).

The number of ways to choose 1 more letter is given by: \(\binom{7}{1} = 7\).

4. Case 4: 3 Es and 2 Rs

We have E, E, E, R, R, and do not need to choose any more letters.

There is only 1 way to choose 0 more letters: \(\binom{6}{0} = 1\).

Adding all the cases together, we get:

\(15 + 6 + 6 + 1 = 28\).

Thus, the total number of possible selections is 28.

The word TOMORROW consists of the letters T, O, M, O, R, R, O, W.

We need to find the probability that a selection of 4 letters contains at least one O and at least one R.

Method 1: Identified scenarios

• OORR: \(\binom{3}{2} \times \binom{2}{2} \times \binom{3}{0} = 3 \times 1 = 3\)
• ORR_: \(\binom{3}{1} \times \binom{2}{2} \times \binom{3}{1} = 3 \times 1 \times 3 = 9\)
• OOR_: \(\binom{3}{2} \times \binom{2}{1} \times \binom{3}{1} = 3 \times 2 \times 3 = 18\)
• OR__: \(\binom{3}{1} \times \binom{2}{1} \times \binom{3}{2} = 3 \times 2 \times 3 = 18\)
• OOOR: \(\binom{3}{3} \times \binom{2}{1} \times \binom{3}{0} = 1 \times 2 = 2\)

\(Total = 50\)

Probability = \(\frac{50}{\binom{8}{4}} = \frac{50}{70}\)

Method 2: Identified outcomes

• ORTM: \(\binom{3}{1} \times \binom{2}{1} = 6\)
• ORTW: \(\binom{3}{1} \times \binom{2}{1} = 6\)
• ORMW: \(\binom{3}{1} \times \binom{2}{1} = 6\)
• ORRM: \(\binom{3}{1} \times \binom{2}{2} = 3\)
• ORRW: \(\binom{3}{1} \times \binom{2}{2} = 3\)
• ORRT: \(\binom{3}{1} \times \binom{2}{2} = 3\)
• OROR: \(\binom{3}{2} \times \binom{2}{1} = 6\)
• OROT: \(\binom{3}{2} \times \binom{2}{1} = 6\)
• OROM: \(\binom{3}{2} \times \binom{2}{1} = 6\)
• OROW: \(\binom{3}{2} \times \binom{2}{1} = 6\)
• OROO: \(\binom{3}{2} \times \binom{2}{1} = 6\)

\(Total = 50\)

Probability = \(\frac{50}{\binom{8}{4}} = \frac{50}{70}\)