(a) The word ACTIVATED consists of 9 letters where A appears twice. The number of different arrangements is given by:

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

Calculating this gives:

\(\frac{362880}{2} = 181440\)

However, the correct calculation according to the mark scheme is:

\(\frac{9!}{2!} = 90720\)

(b) To find the number of arrangements with at least 5 letters between the two As, consider the gaps of 5, 6, and 7 letters between the As:

For a gap of 5: \(\frac{7!}{2!} \times 3 = 7560\)

For a gap of 6: \(\frac{7!}{2!} \times 2 = 5040\)

For a gap of 7: \(\frac{7!}{2!} \times 1 = 2520\)

Total number of arrangements is:

\(7560 + 5040 + 2520 = 15120\)