(i) Consider the three Os as a single unit and the two Ts as another unit. This reduces the problem to arranging the units: \(\{OOO, TT, A, D, S, L\}\), which is 6 units in total. The number of arrangements is \(6! = 720\).

(ii) First, calculate the total number of arrangements of the letters in TOADSTOOL. There are 9 letters with 2 Ts and 3 Os, so the total arrangements are \(\frac{9!}{2!3!} = 30,240\). Next, calculate the number of arrangements where the Ts are together. Treat the Ts as a single unit, reducing the problem to arranging \(\{TT, O, O, O, A, D, S, L\}\), which is 8 units. The number of arrangements is \(\frac{8!}{3!} = 6,720\). Therefore, the number of arrangements where the Ts are not together is \(30,240 - 6,720 = 23,520\).

(iii) To find the probability that a randomly chosen arrangement has a T at the beginning and a T at the end, consider the remaining 7 letters \(\{O, O, O, A, D, S, L\}\). The number of arrangements of these letters is \(\frac{7!}{3!} = 840\). The probability is then \(\frac{840}{30,240} = \frac{1}{36}\) or 0.0278.

(a) First, calculate the total number of arrangements of the letters in CASABLANCA. The word has 10 letters with the following repetitions: A appears 3 times, C appears 2 times. The total arrangements are given by:

\(\frac{10!}{2!4!} = 75600\)

Next, calculate the number of arrangements where the two Cs are together. Treat the two Cs as a single unit, reducing the problem to arranging 9 units (CC, A, S, A, B, L, A, N, A). The arrangements are:

\(\frac{9!}{4!} = 15120\)

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

\(75600 - 15120 = 60480\)

(b) Fix A at the beginning and A at the end, leaving 8 positions to fill. The sequence is A _ _ _ C _ _ _ C _ A. The three letters between the Cs can be chosen from the remaining 6 letters (S, A, B, L, N, A). The number of ways to choose and arrange these 3 letters is:

\(\frac{6!}{2!} \times 4 = 1440\)

(i) Treat the four Es as a single unit. This gives us 9 units to arrange: S, T, EEEE, P, L, C, H, A, S. The number of ways to arrange these 9 units is \(\frac{9!}{2!}\) because there are 2 indistinguishable Ss. Therefore, the number of ways is \(\frac{9!}{2!} = 181,440\).

(ii) First, find the total number of arrangements of the letters in STEEPLECHASE. There are 12 letters with 4 Es and 2 Ss, so the total number of arrangements is \(\frac{12!}{2!4!} = 9,979,200\).

Next, find the number of arrangements where the Ss are together. Treat the two Ss as a single unit, giving us 11 units to arrange: SS, T, E, E, E, E, P, L, C, H, A. The number of ways to arrange these 11 units is \(\frac{11!}{4!} = 1,663,200\).

Finally, subtract the number of arrangements where the Ss are together from the total number of arrangements to find the number of arrangements where the Ss are not together: \(9,979,200 - 1,663,200 = 8,316,000\).

(i) Consider the odd digits (3, 7, 7, 7) as one block and the even digits (2, 2, 8) as another block. The arrangement of these two blocks can be done in 2! ways.

Within the odd block, the digits can be arranged as \(\frac{4!}{3!}\) because there are three 7s.

Within the even block, the digits can be arranged as \(\frac{3!}{2!}\) because there are two 2s.

Therefore, the total number of arrangements is \(2! \times \frac{4!}{3!} \times \frac{3!}{2!} = 2 \times 4 \times 3 = 24\).

(ii) First, calculate the total number of arrangements of the digits without any restriction: \(\frac{7!}{2!3!} = 420\).

Next, calculate the number of arrangements where the two 2s are together. Treat the two 2s as a single unit, so we have 6 units to arrange: \(\frac{6!}{3!} = 120\).

The number of arrangements where the 2s are not together is the total arrangements minus the arrangements where the 2s are together: \(420 - 120 = 300\).

(i) The word CAMERAMAN consists of 9 letters where A appears 3 times and M appears 2 times. The number of arrangements without restrictions is given by:

\(\frac{9!}{3! \times 2!} = \frac{362880}{6 \times 2} = 30240\)

(ii) If the As occupy the 1st, 5th, and 9th positions, we arrange the remaining 6 letters (C, M, E, R, M, N). The number of arrangements is:

\(\frac{6!}{2!} = \frac{720}{2} = 360\)

(iii) If there is exactly one letter between the Ms, consider the block M_X_M where X is any letter. We have 7 positions for this block among the remaining letters (A, A, A, C, E, R, N). The number of arrangements is:

Method 1: Arrange the 7 letters (excluding M) and multiply by 7 positions for the block:

\(\frac{7!}{3!} \times 7 = \frac{5040}{6} \times 7 = 5880\)

Method 2: Choose a letter between Ms and arrange:

\(\frac{6!}{2!} \times 7 + \frac{6!}{3!} \times 7 = 2520 + 3360 = 5880\)