(a) Consider the 3 Es as a single unit. This gives us the units: S, H, O, P, K, (EEE), P, R. There are 8 units in total. The number of ways to arrange these 8 units is \(\frac{8!}{2!}\) because the Ps are repeated. Therefore, the number of ways is \(\frac{8!}{2!} = 20160\).

(b) The total number of ways to arrange the letters of SHOPKEEPER is \(\frac{10!}{2!3!} = 302400\). If the Ps are together, treat them as a single unit: S, H, O, (PP), K, E, E, E, R. This gives us 9 units. The number of ways to arrange these 9 units is \(\frac{9!}{3!} = 60480\). Therefore, the number of ways the Ps are not together is \(302400 - 60480 = 241920\).

(c) The number of ways to have Es at the beginning and end is the number of ways to arrange the remaining 8 letters: S, H, O, P, K, E, P, R. This is \(\frac{8!}{2!} = 20160\). The total number of arrangements is \(\frac{10!}{2!3!} = 302400\). Therefore, the probability is \(\frac{20160}{302400} = \frac{1}{15}\).

(a) The word CELESTIAL has 9 letters with 2 Es. The number of different arrangements is given by:

\(\frac{9!}{2!} = 90,720\)

(b) Fixing C as the first letter, T as the fifth letter, and E as the last letter, we arrange the remaining 6 letters (E, L, S, I, A, L). The number of arrangements is:

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

(c) First, calculate the number of arrangements where the two Es are together. Treat the two Es as a single unit, reducing the problem to arranging 8 units (EE, C, L, S, T, I, A, L):

\(\frac{8!}{2!} = 20,160\)

The number of arrangements where the Es are not together is:

\(90,720 - 20,160 = 70,560\)

The probability that the Es are not together is:

\(\frac{70,560}{90,720} = \frac{7}{9} \approx 0.778\)

(a) To arrange the letters of SUMMERTIME with an E at the beginning and an E at the end, we fix the Es in these positions. This leaves us with 8 letters: S, U, M, M, R, T, I, M. The number of ways to arrange these 8 letters is given by:

\(\frac{8!}{3!}\)

where 3! accounts for the repetition of the letter M. Calculating this gives:

\(\frac{8!}{3!} = \frac{40320}{6} = 6720\)

(b) To find the number of ways to arrange the letters of SUMMERTIME such that the Es are not together, we first find the total number of arrangements of the letters and then subtract the number of arrangements where the Es are together.

The total number of arrangements of the letters is:

\(\frac{10!}{2!3!}\)

where 2! accounts for the repetition of E and 3! for M. Calculating this gives:

\(\frac{10!}{2!3!} = \frac{3628800}{12} = 302400\)

Next, consider the Es as a single unit. This gives us 9 units to arrange: (EE), S, U, M, M, R, T, I, M. The number of ways to arrange these is:

\(\frac{9!}{3!}\)

Calculating this gives:

\(\frac{9!}{3!} = \frac{362880}{6} = 60480\)

Therefore, the number of arrangements where the Es are not together is:

\(302400 - 60480 = 241920\)

(a) Treat the three Es as a single unit and the two Ls as another single unit. This reduces the problem to arranging 6 units: {EEE, LL, J, W, R, Y}. The number of arrangements is given by:

\(\frac{6!}{1!1!1!1!1!1!} = 6! = 720\)

(b) First, calculate the total number of arrangements of the letters in JEWELLERY without any restrictions:

\(\frac{9!}{3!2!} = 30,240\)

Next, calculate the number of arrangements where the two Ls are together. Treat the two Ls as a single unit, reducing the problem to arranging 8 units: {LL, J, E, E, E, W, R, Y}. The number of arrangements is:

\(\frac{8!}{3!} = 6,720\)

Subtract the number of arrangements with the Ls together from the total arrangements to find the number of arrangements where the Ls are not together:

\(30,240 - 6,720 = 23,520\)

(i) The word CORRIDORS has 9 letters where R appears twice. The number of different arrangements is given by:

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

(ii) For the arrangements where the first letter is D and the last letter is R:

We fix D at the start and R at the end, leaving 7 letters: O, R, I, D, O, R, S.

The number of arrangements is:

\(\frac{7!}{2!} = \frac{5040}{2} = 2520\)

For the arrangements where the first letter is D and the last letter is O:

We fix D at the start and O at the end, leaving 7 letters: C, O, R, R, I, D, R.

The number of arrangements is:

\(\frac{7!}{3!} = \frac{5040}{6} = 840\)

\(Total arrangements = 1260 + 840 = 2100.\)