(iii) To have more males than females in a group of 5, we can have the following combinations:

1. 3 males and 2 females: The number of ways to choose 3 males from 5 is \(\binom{5}{3}\), and the number of ways to choose 2 females from 10 is \(\binom{10}{2}\). Thus, the total number of ways is \(\binom{5}{3} \times \binom{10}{2} = 450\).
2. 4 males and 1 female: The number of ways to choose 4 males from 5 is \(\binom{5}{4}\), and the number of ways to choose 1 female from 10 is \(\binom{10}{1}\). Thus, the total number of ways is \(\binom{5}{4} \times \binom{10}{1} = 50\).
3. 5 males and 0 females: The number of ways to choose 5 males from 5 is \(\binom{5}{5}\), and the number of ways to choose 0 females from 10 is \(\binom{10}{0}\). Thus, the total number of ways is \(\binom{5}{5} \times \binom{10}{0} = 1\).

Adding these, the total number of ways is \(450 + 50 + 1 = 501\).

(iv) Mr and Mrs Blake are already chosen, so we need to choose 3 more performers. We need more males than females, so the combinations are:

1. 3 males and 0 females: Choose 3 males from the remaining 4 males. The number of ways is \(\binom{4}{3} \times \binom{9}{0} = 4\).
2. 2 males and 1 female: Choose 2 males from the remaining 4 males and 1 female from 9 females. The number of ways is \(\binom{4}{2} \times \binom{9}{1} = 54\).

Adding these, the total number of ways is \(4 + 54 = 58\).

To solve this problem, we need to consider the different combinations of students that include at least one Chinese and one European student. We will use combinations to calculate the number of ways to choose students.

Let C, E, and A represent the number of Chinese, European, and American students chosen, respectively. We have the following cases:

1. C = 1, E = 1, A = 2: \(7 \times 6 \times \binom{4}{2} = 252\)

2. C = 1, E = 2, A = 1: \(7 \times \binom{6}{2} \times 4 = 420\)

3. C = 1, E = 3, A = 0: \(7 \times \binom{6}{3} \times 1 = 140\)

4. C = 2, E = 1, A = 1: \(\binom{7}{2} \times 6 \times 4 = 504\)

5. C = 2, E = 2, A = 0: \(\binom{7}{2} \times \binom{6}{2} \times 1 = 315\)

6. C = 3, E = 1, A = 0: \(\binom{7}{3} \times 6 \times 1 = 210\)

Adding these cases together gives the total number of ways:

\(252 + 420 + 140 + 504 + 315 + 210 = 1841\)

Since Jon and Sarah must travel in the same taxi, we consider them as a single unit. This leaves us with 6 other friends to arrange.

First, choose 2 more friends to join Jon and Sarah in the same taxi. This can be done in \\(^6C_2\\) ways.

The remaining 4 friends automatically go into the other taxi. Thus, the number of ways to choose the groups is \\(^6C_2\\).

However, Jon and Sarah can be in either taxi P or taxi Q, so we multiply by 2 to account for both possibilities.

Therefore, the total number of ways is \\(^6C_2 \\times 2 = 15 \\times 2 = 30\\).

The total number of people is 16 (7 men + 9 women). We need to find the probability that all 6 people chosen are women.

The number of ways to choose 6 women from 9 is given by the combination formula: \(\binom{9}{6}\).

The total number of ways to choose 6 people from 16 is given by: \(\binom{16}{6}\).

The probability that there are no men on the committee is:

\(P(\text{no men}) = \frac{\binom{9}{6}}{\binom{16}{6}}\)

Calculating these values:

\(\binom{9}{6} = 84\)

\(\binom{16}{6} = 8008\)

Thus, \(P(\text{no men}) = \frac{84}{8008} = \frac{21}{2002} = \frac{3}{286} \approx 0.0105\)

Alternatively, using the probability method:

\(P(\text{no men}) = \frac{9}{16} \times \frac{8}{15} \times \frac{7}{14} \times \frac{6}{13} \times \frac{5}{12} \times \frac{4}{11} = 0.0105\)

To solve this problem, we need to consider all possible combinations where Cherry chooses at least one of each type of ornament. The types are wooden animals (W), sea-shells (S), and pottery ducks (D).

We can break down the combinations as follows:

1. 1 W, 1 S, 3 D: The number of ways to choose 1 wooden animal is 6, 1 sea-shell is 4, and 3 pottery ducks is \(\binom{3}{3} = 1\). Total = \(6 \times 4 \times 1 = 24\).
2. 1 W, 3 S, 1 D: The number of ways to choose 1 wooden animal is 6, 3 sea-shells is \(\binom{4}{3} = 4\), and 1 pottery duck is 3. Total = \(6 \times 4 \times 3 = 72\).
3. 3 W, 1 S, 1 D: The number of ways to choose 3 wooden animals is \(\binom{6}{3} = 20\), 1 sea-shell is 4, and 1 pottery duck is 3. Total = \(20 \times 4 \times 3 = 240\).
4. 1 W, 2 S, 2 D: The number of ways to choose 1 wooden animal is 6, 2 sea-shells is \(\binom{4}{2} = 6\), and 2 pottery ducks is \(\binom{3}{2} = 3\). Total = \(6 \times 6 \times 3 = 108\).
5. 2 W, 1 S, 2 D: The number of ways to choose 2 wooden animals is \(\binom{6}{2} = 15\), 1 sea-shell is 4, and 2 pottery ducks is \(\binom{3}{2} = 3\). Total = \(15 \times 4 \times 3 = 180\).
6. 2 W, 2 S, 1 D: The number of ways to choose 2 wooden animals is \(\binom{6}{2} = 15\), 2 sea-shells is \(\binom{4}{2} = 6\), and 1 pottery duck is 3. Total = \(15 \times 6 \times 3 = 270\).

Adding all these combinations gives the total number of selections:

\(24 + 72 + 240 + 108 + 180 + 270 = 894\).