(a) Consider the even digits 2, 2, 4, 6, 6, 6 as a single block. This block can be arranged in \(\frac{6!}{2!3!}\) ways due to the repeated digits. The odd digits 1, 3, 3 can be arranged in \(\frac{3!}{2!}\) ways. Therefore, the total number of arrangements is \(\frac{6!}{2!3!} \times \frac{3!}{2!} = 720\) ways.

(b) If the first and last digits are both odd, we can have either 1 or 3 at the ends. Consider the cases:

• 1 at the start and 3 at the end: The remaining digits are 2, 2, 3, 4, 6, 6, 6, which can be arranged in \(\frac{7!}{3!2!} = 420\) ways.
• 3 at the start and 1 at the end: The remaining digits are 2, 2, 3, 4, 6, 6, 6, which can be arranged in \(\frac{7!}{3!2!} = 420\) ways.
• 3 at both ends: The remaining digits are 1, 2, 2, 4, 6, 6, 6, which can be arranged in \(\frac{7!}{3!2!} = 420\) ways.

Adding these, the total number of arrangements is \(420 + 420 + 420 = 1260\) ways.

(i) To keep the odd digits together, treat them as a single unit. The odd digits are 1, 3, 5, 7, 9. Arrange these 5 digits in a block: 5! ways. The remaining two digits (4 and 8) can be arranged with this block in 3! ways. Thus, the total number of arrangements is:

\(5! \times 3! = 120 \times 6 = 720\)

(ii) For even numbers between 3000 and 5000, the first digit must be 3 or 4, and the last digit must be 4 or 8. Consider two cases:

• Case 1: Start with 3 and end with 4. The middle two digits can be any of the remaining 5 digits. Thus, there are:

\(5 \times 4 = 20\)

• Case 2: Start with 3 and end with 8. The middle two digits can be any of the remaining 5 digits. Thus, there are:

\(5 \times 4 = 20\)

• Case 3: Start with 4 and end with 8. The middle two digits can be any of the remaining 5 digits. Thus, there are:

\(5 \times 4 = 20\)

Adding these cases gives:

\(20 + 20 + 20 = 60\)

(iii) For multiples of 5 less than 1000, the number must end in 5. Consider numbers with 1, 2, or 3 digits:

• 1-digit: Only 5 is possible.
• 2-digit: The first digit can be any of 1, 3, 4, 7, 8, 9 (6 choices).
• 3-digit: The first two digits can be any combination of the remaining 6 digits (6 choices for the first, 5 for the second).

Thus, the total is:

\(1 + 6 + 6 \times 5 = 1 + 6 + 30 = 57\)

(i) Treat all the A's as a single unit. This gives us the units: {AAA, T, N, Z, N, I}. There are 6 units in total. The number of ways to arrange these units is given by:

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

The division by \(2!\) accounts for the repetition of the letter N.

(ii) We need to arrange the letters such that consonants and vowels alternate, starting with a consonant. The consonants are {T, N, Z, N} and the vowels are {A, A, I}. The number of ways to arrange the consonants is:

\(\frac{4!}{2!}\)

The number of ways to arrange the vowels is:

\(\frac{4!}{3!}\)

Thus, the total number of arrangements is:

\(\frac{4!}{2!} \times \frac{4!}{3!} = 48\)

(i) The word AGGREGATE has 9 letters with the following repetitions: G appears 2 times, A appears 2 times, and E appears 2 times. If the first letter is R, we arrange the remaining 8 letters. The number of arrangements is given by:

\(\frac{8!}{2! \times 2! \times 2!} = \frac{40320}{8} = 1680\)

(ii) Treat the 3 G's as a single unit, the 2 A's as a single unit, and the 2 E's as a single unit. This gives us 5 units to arrange: G, A, E, R, T. The number of arrangements is:

\(5! = 120\)

(iii) We need to arrange the letters such that consonants and vowels alternate. There are 5 consonants (G, G, R, T) and 4 vowels (A, A, E, E). Start with a consonant, then alternate:

\(Consonant positions: 5! / (2!) = 60\)

\(Vowel positions: 4! / (2!) = 12\)

Total arrangements: \(5! \times 4! / (2! \times 2!) = 120\)

(i) The word REMEMBRANCE has 11 letters with repetitions: 3 E's and 2 M's. The number of different arrangements is given by:

\(\frac{11!}{3! \times 2!} = 1663200\)

(ii) For arrangements starting and ending with M, consider the arrangement as M _ _ _ _ _ _ _ _ M. We arrange the remaining 9 letters (REEMBRANC) with 3 E's:

\(\frac{9!}{3!} = 30240\)

(iii) To find arrangements where the 4 vowels (E, E, A, E) are not together, first calculate the arrangements where they are together. Treat the 4 vowels as a single unit, so we have 8 units to arrange (4 vowels + 7 other letters):

\(\frac{8!}{2!} \times \frac{4!}{3!} = 40320\)

Subtract this from the total arrangements:

\(1663200 - 40320 = 1622880\)