(i) To have more girls than boys in the team of 5, the possible combinations are:

• 5 girls and 0 boys: \(\binom{8}{5} \times \binom{6}{0} = 56\)
• 4 girls and 1 boy: \(\binom{8}{4} \times \binom{6}{1} = 420\)
• 3 girls and 2 boys: \(\binom{8}{3} \times \binom{6}{2} = 840\)

\(Total = 56 + 420 + 840 = 1316\)

(ii) If three of the boys are cousins and are either all in the team or all not in the team, consider two cases:

• All cousins in the team: Choose 2 more from the remaining 11 (3 boys + 8 girls): \(\binom{11}{2} = 55\)
• All cousins not in the team: Choose 5 from the remaining 11 (3 boys + 8 girls): \(\binom{11}{5} = 462\)

\(Total = 55 + 462 = 517\)

To find the number of ways to choose the group, we use combinations for each voice part:

1. Choose 10 sopranos from 13: \(\binom{13}{10}\)

2. Choose 9 altos from 12: \(\binom{12}{9}\)

3. Choose 4 tenors from 6: \(\binom{6}{4}\)

4. Choose 4 basses from 7: \(\binom{7}{4}\)

The total number of ways to choose the group is the product of these combinations:

\(\binom{13}{10} \times \binom{12}{9} \times \binom{6}{4} \times \binom{7}{4}\)

Calculating each combination:

\(\binom{13}{10} = 286\)

\(\binom{12}{9} = 220\)

\(\binom{6}{4} = 15\)

\(\binom{7}{4} = 35\)

Thus, the total number of ways is:

\(286 \times 220 \times 15 \times 35 = 33,033,000\)

To select the houses, we need to choose:

1. 1 house from the 2 houses in style B: \(\binom{2}{1} = 2\) ways.
2. 1 house from the 3 houses in style C: \(\binom{3}{1} = 3\) ways.
3. 2 more houses from the remaining 7 houses (2 in style A, 4 in style D, 1 in style E): \(\binom{7}{2} = 21\) ways.

Thus, the total number of ways to select the houses is:

\(2 \times 3 \times 21 = 126\)

To find the number of different possible selections, we use combinations to select the CDs from each category.

First, select 2 pop music CDs from 6: \\(_6C_2 = \frac{6!}{2!(6-2)!} = 15\\)

Next, select 2 jazz CDs from 3: \\(_3C_2 = \frac{3!}{2!(3-2)!} = 3\\)

Finally, select 1 classical CD from 2: \\(_2C_1 = \frac{2!}{1!(2-1)!} = 2\\)

Multiply the combinations together to find the total number of selections: \\(15 \times 3 \times 2 = 90\\)

The word REFRIGERATOR consists of the letters: R, E, F, R, I, G, E, R, A, T, O, R.

We need to select four letters with no Rs and exactly two Es.

First, remove all Rs from the word, leaving: E, F, I, G, E, A, T, O.

Now, we have 8 letters: E, E, F, I, G, A, T, O.

We need to select 4 letters, including exactly 2 Es.

\(Choose 2 Es: inom{2}{2} = 1.\)

\(Choose 2 more letters from the remaining 6 letters (F, I, G, A, T, O): inom{6}{2} = 15.\)

\(Thus, the total number of selections is: 1 imes 15 = 15.\)