The word ANDROMEDA consists of the letters A, N, D, R, O, M, E, D, A. There are 9 letters in total.

We need to find the probability of selecting 4 letters such that there is at least one D and exactly one A.

Method 1:

Consider the scenarios:

1. Scenario 1 (AD^): Select 1 A, 1 D, and 2 other letters from the remaining 7 letters (excluding the second D). The number of ways to do this is \(\binom{2}{1} \times \binom{2}{1} \times \binom{5}{2} = 40\).
2. Scenario 2 (ADD^): Select 1 A, 2 Ds, and 1 other letter from the remaining 6 letters. The number of ways to do this is \(\binom{2}{1} \times \binom{1}{1} \times \binom{6}{1} = 10\).

\(Total number of favorable selections = 40 + 10 = 50.\)

Total number of ways to select 4 letters from 9 = \(\binom{9}{4} = 126\).

Thus, the probability is \(\frac{50}{126} = \frac{25}{63}\).

We need to select 6 letters from the word CATERPILLAR, which contains the letters C, A, T, E, R, P, I, L, L, A, R. The selection must include both Rs, at least one A, and at least one L.

Method 1:

1. Fix R, R, A, L. We need to choose 2 more letters from the remaining 5 letters: C, T, E, P, I.
2. Calculate the combinations: \\(^5C_2 = 10\).
3. Fix R, R, A, L, L. Choose 1 more letter from C, T, E, P, I: \\(^5C_1 = 5\).
4. Fix R, R, A, A, L. Choose 1 more letter from C, T, E, P, I: \\(^5C_1 = 5\).
5. Fix R, R, A, A, L, L. No more letters needed: 1 way.

\(Total = 10 + 5 + 5 + 1 = 21.\)

Method 2:

1. Fix R, R, A, L. Choose 2 more letters from the remaining 7 letters: C, T, E, P, I, A, L.
2. Calculate the combinations: \\(^7C_2 = 21\).

Both methods give the total number of selections as 21.

The word SHOPKEEPER consists of the letters: S, H, O, P, K, E, E, P, E, R.

We need to select four letters including exactly one P. The possible scenarios are:

1. P E E E: Choose 3 E's from the 3 available E's. This can be done in \(\binom{3}{3} = 1\) way.
2. P E E _: Choose 2 E's from the 3 available E's and 1 other letter from the remaining 5 distinct letters (S, H, O, K, R). This can be done in \(\binom{3}{2} \times \binom{5}{1} = 3 \times 5 = 15\) ways.
3. P E _ _: Choose 1 E from the 3 available E's and 2 other letters from the remaining 5 distinct letters. This can be done in \(\binom{3}{1} \times \binom{5}{2} = 3 \times 10 = 30\) ways.
4. P _ _ _: Choose 3 other letters from the remaining 5 distinct letters. This can be done in \(\binom{5}{3} = 10\) ways.

Summing these possibilities gives the total number of selections:

\(1 + 15 + 30 + 10 = 56\)

However, according to the mark scheme, the correct total is 26. This suggests a different interpretation or constraint not visible in the question text.

The word 'CELESTIAL' consists of the letters: C, E, L, E, S, T, I, A, L.

We need to select 5 letters such that there is at least one E and at most one L.

Consider the scenarios:

1. One E and one L: Choose 3 more letters from the remaining 5 letters (C, S, T, I, A). This can be done in \(\binom{5}{3} = 10\) ways.
2. Two Es and one L: Choose 2 more letters from the remaining 5 letters (C, S, T, I, A). This can be done in \(\binom{5}{2} = 10\) ways.
3. One E and no L: Choose 4 more letters from the remaining 6 letters (C, S, T, I, A, L). This can be done in \(\binom{5}{4} = 5\) ways.
4. Two Es and no L: Choose 3 more letters from the remaining 6 letters (C, S, T, I, A, L). This can be done in \(\binom{5}{3} = 10\) ways.

Summing the number of ways for these scenarios gives:

\(10 + 10 + 5 + 10 = 35\)

The word SUMMERTIME consists of the letters: S, U, M, M, E, R, T, I, M, E.

We need to select 4 letters including at least one M and exactly one E.

Consider the scenarios:

1. Scenario 1: EMMM

We have already selected 3 M's and 1 E. The number of ways to choose 0 more letters from the remaining 5 letters (S, U, R, T, I) is:

\(\binom{5}{0} = 1\)

2. Scenario 2: EMM_

We have selected 2 M's and 1 E. We need to choose 1 more letter from the remaining 5 letters (S, U, R, T, I):

\(\binom{5}{1} = 5\)

3. Scenario 3: EM__

We have selected 1 M and 1 E. We need to choose 2 more letters from the remaining 5 letters (S, U, R, T, I):

\(\binom{5}{2} = 10\)

Summing the number of ways for the scenarios gives:

\(1 + 5 + 10 = 16\)