(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:

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