1. ## Combinations Help

Ive been trying to do these questions for hours, cant think of a solution. Help would be appreciated !

1) In how many ways could 15 different books be divided equally among 3 people? Answer= 756756

2) the camer club has 5 members and the math club has 8. there is only 1 member common to both clubs. In how many ways could a committee of 4 people be formed with atleast one member from each club? Answer=459

2. 2) the camer club has 5 members and the math club has 8. there is only 1 member common to both clubs. In how many ways could a committee of 4 people be formed with atleast one member from each club? Answer=459
Since we want at least one member from each, the best thing to do is to find no memebrs from each and subtract from the total number of possibilities.

Since there is a common member we must account for that.

The total number of choices is C(12,4)=495

Just math members would be C(7,4)=35

Just camer members would be C(4,4)=1

495-35-1=459

3. Hello, iaimforperfect!

1) In how many ways could 15 different books
be divided equally among 3 people? . Answer: 756756
This is a "partition" problem.

The answer is: . ${15\choose5,5,5} \:=\:\frac{15!}{5!5!5!} \;=\;756,756$