- #1

nowimpsbball

- 15

- 0

## Homework Statement

Suppose 30 identical juggling balls are distributed to 5 different jugglers.

A) how many ways can the balls be distributed so that each juggler receives at least three balls?

B)In how many ways can the balls be distributed so that each juggler receives between 3 and 7 balls?

## The Attempt at a Solution

A) I understand perfectly, the ans is C(19,4)...I get that

B) The answer is C(19,4)-5C(14,4)+C(5,2)C(9,4)-C(5,3)C(4,4) = 121

From the book "the number of distributions can be represented by solutions of the linear equation x1+x2+x3+x4+x5=30. In this case, each xi satisfies 3<=xi<=7. We proceed with the help of the principle of inclusion-exclusion . First consider all distributions with xi>=3, C(19,4). If Pi is the property that xi>=8, we count only those distributions that satisfy none of the properties Pi. The Total count is (19,4)-5

**C(14,4)**+C(5,2)

**C(9,4)**-C(5,3)

**C(4,4)**= 121."

I will bold the parts I do not understand. I was sick the day this was went over in class. One last question, why do you not consider the intersect of four and five "jugglers"...other words C(5,2) is the intersect of 2 jugglers, C(5,3) is the intersect of 3 jugglers, at least that is how I perceive it. They must be 0, but why would they be 0 in this case?

Thanks