(a) To arrange the letters such that no consonant is next to another consonant, we first arrange the vowels A, O, E, A. There are 4 vowels, and they can be arranged in \(\frac{4!}{2!}\) ways due to the repetition of A. This gives \(\frac{4!}{2!} = 12\) ways.

Next, we place the 5 consonants (D, M, N, R, D) in the gaps between and around the vowels. There are 5 gaps (before, between, and after the vowels), and we need to choose 4 of these gaps to place the consonants. The consonants can be arranged in \(\frac{5!}{2!}\) ways due to the repetition of D. This gives \(\frac{5!}{2!} = 60\) ways.

The total number of arrangements is \(12 \times 60 = 720\).

(b) To arrange the letters with an A at each end and the Ds not together, we first fix the As at each end: A _ _ _ _ _ _ _ A. This leaves 7 positions for the remaining letters (D, N, D, R, O, M, E).

First, calculate the total arrangements with A at each end: \(\frac{7!}{2!} = 2520\) ways, due to the repetition of D.

Next, calculate the arrangements where the Ds are together. Treat the two Ds as a single unit, so we have (DD, N, R, O, M, E) to arrange, which is \(6! = 720\) ways.

The number of arrangements where the Ds are not together is \(2520 - 720 = 1800\).

(a) The word TOMORROW consists of 8 letters where T, M, and W are unique, O appears 3 times, and R appears 2 times. The total number of different arrangements is given by:

\(\frac{8!}{2!3!} = \frac{40320}{12} = 3360\)

(b) To find the number of arrangements with R at the beginning and end, and the Os not all together, we first consider the arrangement of the remaining 6 letters (T, O, M, O, W, O). The total arrangements of these 6 letters is:

\(\frac{6!}{3!} = 120\)

Next, we subtract the cases where all Os are together. Treating the three Os as a single unit, we arrange T, M, W, and the O unit:

\(4! = 24\)

Thus, the number of arrangements where the Os are not all together is:

\(120 - 24 = 96\)

(a) The word RELEASED consists of 8 letters where E is repeated twice. The number of different arrangements is given by \(\frac{8!}{2!}\). Calculating this gives \(\frac{40320}{2} = 6720\).

(b) Treat LED as a single unit. This gives us 6 units to arrange: LED, R, E, A, S, E. The number of arrangements is \(\frac{6!}{2!}\) because E is repeated. Calculating this gives \(\frac{720}{2} = 360\).

(c) First, calculate the total number of arrangements where A and D are together. Treat AD as a single unit, giving us 7 units: AD, R, E, L, E, S, E. The number of arrangements is \(\frac{7!}{3!}\) because E is repeated. Calculating this gives \(\frac{5040}{6} = 840\). The probability that A and D are not together is \(1 - \frac{840}{6720} = \frac{3}{4}\) or 0.75.

(a) The word CATERPILLAR consists of 11 letters where C, A, T, E, P, I, L, L, A, R, R are present. The letters A, L, and R are repeated. The formula for permutations of letters with repetitions is given by:

\(\frac{11!}{2!2!2!}\)

Calculating this gives:

\(\frac{11!}{2!2!2!} = \frac{39916800}{8} = 4989600\)

(b) To find the number of arrangements with R at the beginning and end, and the two As not together, we use two methods:

Method 1:

Arrange the 7 letters CTEPILL (excluding the two Rs and two As):

\(\frac{7!}{2!}\)

Calculate the number of ways to place As in non-adjacent places:

\(\binom{8}{2}\)

The total number of arrangements is:

\(\frac{7!}{2!} \times \binom{8}{2} = 2520 \times 28 = 70560\)

Method 2:

Total arrangements with R at the beginning and end:

\(\frac{9!}{2!2!}\)

Arrangements with R at ends and As together:

\(\frac{8!}{2!}\)

With As not together:

\(\frac{9!}{2!2!} - \frac{8!}{2!} = 90720 - 20160 = 70560\)

(a) The total number of ways to arrange the letters in RESERVED is \(\frac{8!}{3!2!} = 3360\) because there are 3 Es and 2 Rs.

To have V as the first letter and E as the last letter, we arrange the remaining 6 letters: R, E, S, E, R, D. The number of ways to arrange these is \(\frac{6!}{2!2!} = 180\).

The probability is \(\frac{180}{3360} = \frac{3}{56}\) or 0.0536.

(b) Consider Rs together and Es together as single units: [RR] and [EEE]. This leaves us with the units [RR], [EEE], S, V, D to arrange, which is 5! ways.

The number of ways to arrange the Es together is \(\frac{6!}{2!} = 360\).

The probability is \(\frac{5!}{\frac{6!}{2!}} = \frac{120}{360} = \frac{1}{3}\).