(c) To divide 9 members into a group of 5 and a group of 4, we choose 5 members out of 9. This can be calculated using combinations:

\(\binom{9}{5} = 126\)

(d) To find the probability that Raman and Sanjay are in the same group, consider two cases:

1. Both are in the group of 5. Choose 3 more members from the remaining 7:

\(\binom{7}{3} = 35\)

2. Both are in the group of 4. Choose 2 more members from the remaining 7:

\(\binom{7}{2} = 21\)

Total ways for both to be in the same group:

\(35 + 21 = 56\)

Probability that Raman and Sanjay are in the same group:

\(\frac{56}{126} = \frac{4}{9} \approx 0.444\)

To find the number of different selections of 5 marbles containing at least 4 marbles of the same color, we consider two cases:

1. 4 red marbles and 1 blue marble:

   The number of ways to choose 4 red marbles from 8 is given by \(\binom{8}{4}\), and the number of ways to choose 1 blue marble from 4 is \(\binom{4}{1}\).

   Thus, the total number of ways for this case is \(\binom{8}{4} \times \binom{4}{1} = 70 \times 4 = 280\).

2. 4 blue marbles and 1 red marble:

   The number of ways to choose 4 blue marbles from 4 is \(\binom{4}{4}\), and the number of ways to choose 1 red marble from 8 is \(\binom{8}{1}\).

   Thus, the total number of ways for this case is \(\binom{4}{4} \times \binom{8}{1} = 1 \times 8 = 8\).

3. 5 red marbles:

   The number of ways to choose 5 red marbles from 8 is \(\binom{8}{5}\).

   Thus, the total number of ways for this case is \(\binom{8}{5} = 56\).

Adding all these cases together, the total number of selections is:

\(280 + 8 + 56 = 344\).

(a) To have more women than men, consider the following scenarios:

1. 6 women and 0 men: \(\binom{9}{6} = 84\)
2. 5 women and 1 man: \(\binom{9}{5} \times \binom{5}{1} = 126 \times 5 = 630\)
3. 4 women and 2 men: \(\binom{9}{4} \times \binom{5}{2} = 126 \times 10 = 1260\)

\(Total number of ways = 84 + 630 + 1260 = 1974.\)

(b) Total number of ways to choose 6 people from 14: \(\binom{14}{6} = 3003\).

Number of ways with both sister and brother: \(\binom{12}{4} = 495\).

Number of ways without both sister and brother: \(3003 - 495 = 2508\).

Alternatively, calculate:

1. Number of ways with neither: \(\binom{12}{6} = 924\)
2. Number of ways with either brother or sister (not both): \(\binom{12}{5} \times 2 = 792 \times 2 = 1584\)

\(Total number of ways = 924 + 1584 = 2508.\)

(a) To divide 9 people into a group of 6 and a group of 3, we calculate the number of combinations of choosing 6 people out of 9. This is given by:

\(\binom{9}{6} = \binom{9}{3} = 84\)

Thus, there are 84 ways to divide the 9 people into a group of 6 and a group of 3.

(b) To find the probability that a group of 5 people contains all 3 of the Baker children, we first calculate the number of ways to choose 3 Baker children and 2 others from the remaining 6 people (4 Ahmeds and 2 Bakers):

Number of ways to choose 2 from 6: \(\binom{6}{2} = 15\)

Total number of ways to choose any 5 people from 9: \(\binom{9}{5} = 126\)

The probability is then:

\(\frac{15}{126} = \frac{5}{42}\)

We need to select 7 musicians with at least 2 pianists, at least 1 guitarist, and more violinists than guitarists. Let's consider the possible scenarios:

1. 2 Pianists, 3 Violinists, 2 Guitarists:

   \(\binom{8}{2} \times \binom{4}{2} \times \binom{6}{3} = 28 \times 6 \times 20 = 3360\)

2. 2 Pianists, 4 Violinists, 1 Guitarist:

   \(\binom{8}{2} \times \binom{4}{1} \times \binom{6}{4} = 28 \times 4 \times 15 = 1680\)

3. 3 Pianists, 3 Violinists, 1 Guitarist:

   \(\binom{8}{3} \times \binom{4}{1} \times \binom{6}{3} = 56 \times 4 \times 20 = 4480\)

4. 4 Pianists, 2 Violinists, 1 Guitarist:

   \(\binom{8}{4} \times \binom{4}{1} \times \binom{6}{2} = 70 \times 4 \times 15 = 4200\)

Summing the number of ways for all scenarios gives:

\(3360 + 1680 + 4480 + 4200 = 13,720\)