Hi, Lj.

1. Part One: Considering that all boys and all girls must be side-by-side, we will end up with one group of 4 boys, and one group of 6 girls.

There are possible ways to arrange the group of four boys. Similarly, there are ways to arrange the group of six girls.

These groups may be arranged in 2 ways: boys on the right, girls on the left, or vice-versa, making a total of:

possible arrangements.

Part Two: If only the girls need to be grouped together, then consider that the group of girls may not only be placed to the right or left of the boys, but also "inserted" between any two boys. This makes a total of 5 positions for placing the group of girls. So, there are:

possible arrangements.

2. If I'm not misunderstanding the problem statement, then perhaps it contains a small error: If the 17 students are to be divided into

... that makes 6 committees not 5.

We may place the 17 students into the 6 committees in:

ways.

However, this assumes the two committees of 2 students and the three committees of 3 students are distinguishable; in fact, they are indistinguishable from one another. We have overcounted the 2-student committees 2! times, and the 3-student committees 3! times.

Therefore, the correct number of arrangements is:

arrangements.

3. The probability that the player will sink any particular shot is 0.50.

Out of the player's next 10 shots, choose 3 to be successful in ways.

Then, the probability that exactly 3 of the next 10 shots will be successful is:

, or about 12%.

For Problem No. 2
How about if students are allowed to be a member of more than one committee even allowed to become a member on all committee? I did not make a solution to it, I just raise a situation here because the problem did not specifically specify that the student is allowed to become a member of more than one committee.