The word VENEZUELA consists of the letters V, E, N, E, Z, U, E, L, A. There are three E's in the word.

To select four letters containing exactly one E, we first choose one E. This leaves us with six other letters: V, N, Z, U, L, A.

We need to choose three more letters from these six. The number of ways to choose 3 letters from 6 is given by the combination formula:

\(\binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20\)

Thus, the number of possible selections containing exactly one E is 20.

(a) To form a 3-digit number with no repeated digits, we choose 3 different digits from the available numbers: 1, 2, 3, 4, 6. There are 5 different digits, so we calculate the permutations of choosing 3 out of 5:

\(^5P_3 = \frac{5!}{(5-3)!} = 5 \times 4 \times 3 = 60\)

Thus, there are 60 different numbers that can be made with no repeated digits.

(b) To form a number between 200 and 300, the first digit must be 2. The remaining two digits can be chosen from 1, 2, 3, 3, 4, 6, 6, 6. We list the possible numbers:

212, 213, 214, 216, 221, 223, 224, 226, 231, 232, 233, 234, 236, 241, 242, 243, 246, 261, 262, 263, 264, 266

There are 22 such numbers.

(iii) To select exactly 1 N and 1 A, we choose 1 N and 1 A from the letters TANZANIA. The remaining letters are T, Z, I, and the other A. We need to choose 2 more letters from these 3 letters. This can be done in \(\binom{3}{2} = 3\) ways.

(iv) To select exactly 1 N, we consider different cases based on the number of A's:

1. 0 A's: Select N and 3 other letters from T, Z, I, A. This is 1 way: N, T, Z, I.
2. 2 A's: Select N, 2 A's, and 1 other letter from T, Z, I. This can be done in \(\binom{3}{1} = 3\) ways.
3. 3 A's: Select N and 3 A's. This is 1 way: N, A, A, A.

Adding these, the total number of ways is \(1 + 3 + 1 = 8\) ways.

To solve this problem, we need to consider two cases: selecting 4 letters with exactly 2 Gs and selecting 4 letters with exactly 3 Gs.

Case 1: Exactly 2 Gs

We choose 2 Gs from the 3 available Gs. The remaining 2 letters must be chosen from the letters A, E, R, T. The possible combinations are:

• GG with AA
• GG with AE
• GG with EE
• GG with RA
• GG with RE
• GG with RT
• GG with TA
• GG with TE

This gives us 8 ways.

Case 2: Exactly 3 Gs

We choose 3 Gs from the 3 available Gs. The remaining letter must be chosen from A, E, R, T. The possible combinations are:

• GGG with A
• GGG with E
• GGG with R
• GGG with T

This gives us 4 ways.

Total

\(Adding both cases together, we have a total of 8 + 4 = 12 ways.\)

The word 'REMEMBRANCE' consists of the letters: R, E, M, E, M, B, R, A, N, C, E.

Removing M and R, we have the letters: E, E, B, A, N, C, E.

We need to choose 4 letters with at least 2 Es.

Case 1: Exactly 2 Es. Choose 2 more letters from B, A, N, C.

The number of ways to choose 2 Es from 3 Es is \(\binom{3}{2} = 3\).

The number of ways to choose 2 more letters from B, A, N, C is \(\binom{4}{2} = 6\).

Total for this case: \(3 \times 6 = 18\) ways.

Case 2: Exactly 3 Es. Choose 1 more letter from B, A, N, C.

The number of ways to choose 3 Es from 3 Es is \(\binom{3}{3} = 1\).

The number of ways to choose 1 more letter from B, A, N, C is \(\binom{4}{1} = 4\).

Total for this case: \(1 \times 4 = 4\) ways.

Adding both cases: \(18 + 4 = 22\) ways.

Alternatively, choose 2 letters from E, E, E, B, A, N, C (5 letters): \(\binom{5}{2} = 10\) ways.

Thus, the total number of ways is 10.