(i) Consider all the E's as a single unit. This gives us the units: (EEEE), T, N, N, S, S. We have 6 units in total.

The number of ways to arrange these 6 units is \(\frac{6!}{2!2!}\) because there are 2 N's and 2 S's which are identical.

Calculating this gives \(\frac{720}{4} = 180\).

(ii) Fix T at one end and S at the other end. This leaves us with the letters E, E, E, E, N, N, S to arrange.

The number of ways to arrange these 7 letters is \(\frac{7!}{4!2!}\) because there are 4 E's and 2 N's which are identical.

Calculating this gives \(\frac{5040}{48} = 105\).

Since T can be at either end, we multiply by 2: \(105 \times 2 = 210\).

(i) With the first letter as N and the last letter as B, the arrangement is N****B. We have 5 letters left: A, A, A, N, S. The number of ways to arrange these 5 letters is given by:

\(\frac{5!}{3!} = 20\)

since there are 3 A's which are identical.

(ii) If all the letters A are next to each other, treat them as a single unit (AAA). The letters to arrange are B, (AAA), N, N, S. The number of ways to arrange these 5 units is:

\(\frac{5!}{2!} = 60\)

since there are 2 N's which are identical.

(i) To find the number of different numbers without restrictions, we calculate the permutations of the digits 223 677 888. The total number of digits is 9, with repetitions: two 2s, two 3s, and three 8s. The formula for permutations with repetitions is:

\(\frac{9!}{2!2!3!} = \frac{362880}{8 \times 2 \times 6} = 15120 \text{ ways}\)

(ii) To find the number of even numbers, the last digit must be 2, 6, or 8. We calculate the permutations for each case:

• Ending with 2: Arrange the remaining 8 digits (23677888), with repetitions of two 8s and two 7s:

\(\frac{8!}{2!2!} = \frac{40320}{4 \times 2} = 3360 \text{ ways}\)

• Ending with 6: Arrange the remaining 8 digits (22377888), with repetitions of two 2s, two 8s, and two 7s:

\(\frac{8!}{2!2!2!} = \frac{40320}{8} = 1680 \text{ ways}\)

• Ending with 8: Arrange the remaining 8 digits (22367788), with repetitions of two 2s and two 7s:

\(\frac{8!}{2!2!} = \frac{40320}{4 \times 2} = 5040 \text{ ways}\)

Total even numbers:

\(3360 + 1680 + 5040 = 10080 \text{ ways}\)

(a) Consider the three Es as a single unit, reducing the problem to arranging 7 units: \(\{EEE, D, D, L, I, V, R\}\). The total arrangements of these 7 units is \(\frac{7!}{2!}\) due to the repetition of the two Ds. This gives \(\frac{5040}{2} = 2520\).

Now, subtract the arrangements where the two Ds are together. Treat the two Ds as a single unit, reducing the problem to arranging 6 units: \(\{EEE, DD, L, I, V, R\}\). The total arrangements of these 6 units is \(6! = 720\).

Thus, the number of arrangements where the three Es are together and the two Ds are not next to each other is \(2520 - 720 = 1800\).

(b) The total number of arrangements of the 9 letters is \(\frac{9!}{3!2!} = 362880\).

To find the number of arrangements with exactly 4 letters between the two Ds, consider the positions of the Ds. If the first D is in position 1, the second D can be in position 6. The letters between them can be any 4 of the remaining 7 letters. The number of ways to choose 4 letters from 7 is \(\binom{7}{4} = 35\).

For each choice of 4 letters, there are \(4!\) ways to arrange them. Thus, the number of arrangements for this scenario is \(35 \times 24 = 840\).

Since there are 5 possible starting positions for the first D (1 to 5), the total number of favorable arrangements is \(5 \times 840 = 4200\).

The probability is then \(\frac{4200}{362880} = \frac{1}{9}\).

To find the number of possible arrangements where the sum of the numbers at each end of the line is 4, we consider the possible pairs for the ends of the line: (1, 3), (3, 1), and (2, 2).

For each of these pairs, the remaining 5 dice can show any of the 6 faces. Therefore, there are:

\(6^5\) ways to arrange the middle 5 dice.

Since there are 3 possible pairs for the ends, the total number of arrangements is:

\(6^5 \times 3 = 7776 \times 3 = 23328\)