(i) No digit can be repeated:

To form a 3-digit number greater than 300, the hundreds digit must be 3, 4, 6, or 8. This gives us 4 options for the hundreds digit.

After choosing the hundreds digit, 5 digits remain for the tens place, and 4 digits remain for the units place.

The total number of such numbers is given by:

\(4 \times 5 \times 4 = 80\)

(ii) A digit can be repeated and the number made is even:

The number must be even, so the units digit must be 2, 4, 6, or 8, giving us 4 options for the units digit.

The hundreds digit must be 3, 4, 6, or 8 to ensure the number is greater than 300, giving us 4 options for the hundreds digit.

The tens digit can be any of the 6 digits (1, 2, 3, 4, 6, 8), giving us 6 options for the tens digit.

The total number of such numbers is given by:

\(4 \times 6 \times 4 = 96\)

(i) Digits are not repeated:

The first digit must be 3 or 4 to ensure the number is between 3000 and 5000. This gives us 2 choices for the first digit.

After choosing the first digit, we have 4 remaining digits. We need to choose 3 more digits from these 4, which can be done in:

\(4 \times 3 \times 2 = 24\)

Thus, the total number of ways is:

\(2 \times 24 = 48\)

(ii) Digits can be repeated and the number is odd:

The last digit must be odd, so it can be 1, 3, or 5. This gives us 3 choices for the last digit.

The first digit must be 3 or 4 to ensure the number is between 3000 and 5000. This gives us 2 choices for the first digit.

The middle two digits can be any of the 5 digits (1, 2, 3, 4, 5) and can be repeated. Therefore, there are:

\(5 \times 5 = 25\)

ways to choose the middle two digits.

Thus, the total number of ways is:

\(2 \times 25 \times 3 = 150\)

(a) The word COCOONED has 8 letters with repetitions: 2 Cs, 3 Os. The number of different arrangements is given by:

\(\frac{8!}{2!3!} = 3360\)

(b) Fixing the first letter as O and the last letter as N, we arrange the remaining 6 letters: C, C, O, O, E. The number of arrangements is:

\(\frac{6!}{2!2!} = 180\)

(c) Treat the three Os as a single unit and the two Cs as another unit. We have the units: OOO, CC, E, N, D. The number of arrangements of these 5 units is:

\(\frac{5!}{1!} = 120\)

The number of arrangements where the two Cs are together is:

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

The probability is:

\(\frac{120}{840} = \frac{1}{7} \approx 0.143\)

(i) To form a 3-digit odd number using the digits 4, 5, 6, 7, the last digit must be odd. The odd digits available are 5 and 7.

For numbers ending in 5, the remaining two digits can be chosen from 4, 6, 7. This can be done in \(^3P_2 = 6\) ways.

For numbers ending in 7, the remaining two digits can be chosen from 4, 5, 6. This can also be done in \(^3P_2 = 6\) ways.

Therefore, the total number of 3-digit odd numbers is \(6 + 6 = 12\).

(ii) Consider numbers with different numbers of digits:

• 1-digit numbers: The odd digits are 5 and 7, so there are 2 numbers.
• 2-digit numbers: The last digit must be odd (5 or 7). For each choice of the last digit, the first digit can be chosen from the remaining 3 digits. This gives \(^3P_1 \times 2 = 6\) numbers.
• 3-digit numbers: As calculated in part (i), there are 12 numbers.
• 4-digit numbers: The last digit must be odd (5 or 7). For each choice of the last digit, the first three digits can be chosen from the remaining 3 digits. This gives \(^3P_3 \times 2 = 12\) numbers.

Therefore, the total number of odd numbers is \(2 + 6 + 12 + 12 = 32\).

(i) Consider the vowels (A, E, O) as a single unit and the consonants (C, G, H, N, P) as another unit. This gives us two units: (OAEE) and (CPNHGN). The arrangement of these two units can be done in 2! ways.

Within the vowels, A, E, and O can be arranged in 4!/2! ways because there are two Es.

Within the consonants, C, G, H, N, and P can be arranged in 6!/2! ways because there are two Ns.

Thus, the total number of arrangements is:

\(2! \times \frac{4!}{2!} \times \frac{6!}{2!} = 2 \times 12 \times 360 = 8640\)

(ii) First, calculate the total number of arrangements of the letters in COPENHAGEN without any restrictions. This is given by:

\(\frac{10!}{2!2!} = 907200\)

Next, calculate the number of arrangements where the two Es are together. Treat the two Es as a single unit, reducing the problem to arranging 9 units (including the double E) with the repeated N:

\(\frac{9!}{2!} = 181440\)

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

\(907200 - 181440 = 725760\)

Alternatively, consider arranging the letters C, P, N, H, G, N, O, A in \(\frac{8!}{2!}\) ways, then insert the first E in 9 possible positions and the second E in 8 possible positions, dividing by 2 to account for the indistinguishable Es:

\(\frac{8!}{2!} \times 9 \times 8 \div 2 = 725760\)