(i) Consider each couple as a single unit or 'block'. There are 7 such blocks. These blocks can be arranged in a line in \(7!\) ways. Within each block, the two people can be arranged in \(2!\) ways. Therefore, the total number of arrangements is \(7! \times 2^7 = 645,120\).

(ii) Treat the 7 friends as one group and the 7 partners as another group. These two groups can be arranged in \(2!\) ways. Within each group, the individuals can be arranged in \(7!\) ways. Therefore, the total number of arrangements is \(7! \times 7! \times 2 = 50,803,200\).

To solve this problem, we treat each folder as a single unit or block. Thus, we have 3 blocks: Family, Holiday, and Friends.

The number of ways to arrange these 3 blocks is given by the factorial of the number of blocks:

\(3! = 6\)

Within each block, the images can be arranged among themselves. Therefore, we calculate the factorial for each folder:

Family folder: \(3! = 6\)

Holiday folder: \(4! = 24\)

Friends folder: \(8! = 40,320\)

The total number of arrangements is the product of these factorials:

\(3! \times 3! \times 4! \times 8! = 6 \times 6 \times 24 \times 40,320 = 34,836,480\)

1. Each coin can show either a head or a tail, giving 2 possibilities per coin. For 12 coins, the total number of arrangements is given by:

   \(2^{12} = 4096\)

2. To find the number of arrangements with 7 heads and 5 tails, use the combination formula:

   \(\binom{12}{7} = \frac{12!}{7!5!} = 792\)

(i) When the order of choosing is taken into account, we need to multiply the number of combinations by the number of permutations of the 4 astronauts. The number of permutations of 4 astronauts is given by:

\(4! = 4 \times 3 \times 2 \times 1 = 24\)

Thus, the total number of ways is:

\(3876 \times 24 = 93024\)

(ii) Each astronaut has 3 possessions, and these possessions are kept together. We treat each astronaut's possessions as a single unit, so we have 4 units to arrange. The number of ways to arrange these 4 units is:

\(4! = 24\)

Within each unit, the 3 possessions can be arranged in:

\(3! = 6\) ways

Therefore, the total number of ways to arrange the possessions is:

\((3!)^4 \times 4! = 6^4 \times 24 = 31104\)

(i) To find the number of different outfits, multiply the number of choices for each item: shoes, jeans, and tee shirts. Thus, the total number of outfits is:

\(4 \times 3 \times 7 = 84\)

(ii) First, calculate the total number of arrangements of 10 items (3 jeans + 7 tee shirts):

\(10! = 3628800\)

Next, calculate the arrangements where the two blue items are together. Treat the two blue items as a single unit, reducing the problem to arranging 9 items:

\(9! = 362880\)

Within this unit, the two blue items can be arranged in \(2!\) ways:

\(2! = 2\)

Thus, the number of arrangements where the blue items are together is:

\(9! \times 2 = 725760\)

Subtract this from the total arrangements to find the number of ways the blue items are not together:

\(10! - 9! \times 2 = 2903040\)

(iii) Fahad can choose any number of books from 1 to 9. The number of ways to choose at least one book is the total number of subsets of 9 books minus the empty set:

The total number of subsets is \(2^9\):

\(2^9 = 512\)

Subtract the empty set:

\(512 - 1 = 511\)