Hey there faisal,
Because of time zones, I'm not sure if I'm too late for you...
1. In how many way could 15 different books be divided equally among 3 people?
Answer in the book = 756756
15 books can be divided 15! ways = 1307674368000
Even though the books are different, the order within the group is not important. ie. ABCDE = ADECB (both groups contain the same letters). Each group of 5 books can be divided in 5! ways.
Hence, 15 different books are divided eually among 3 people
2. The camera club has 5 members, and the mathematics club has 8. There is only one member common to both clubs. In how many ways could a committee of 4 people be formed with at least one member from each club?
Answer in the book = 459
If you know what a venn diagram is, it might be helpful to draw problems like this using one...
There are 12 members all together (5 + 8 - 1(the common member) = 12)
These 12 members can be arranged in ways.
This number represents the total permutations for the group. Taking out the other groups which contain only members from one group:
I think you've mixed up some information for your 3rd question
Sorry about the delay - I'm on child-duties at home :P
Trust that helps...