Need help with discrete Advance counting math question

Question

After another gym class you are tasked with putting the 13 identical dodgeballs away into 3 bins. Each bin can hold at most 5 balls. How many ways can you clean up?

Answer 1

To solve this problem, we can use combinatorics, specifically the concept of distributing identical objects into distinct bins with a capacity constraint.

Let's denote the number of dodgeballs asn=13 and the number of bins ask=3. We need to find the number of ways to distribute the dodgeballs into the bins such that each bin can hold at most 5 balls.

To solve this, we can use a technique called stars and bars. Imagine each dodgeball as a star (∗∗) and each bin as a divider (|). For example, if there are 13 dodgeballs and 3 bins, one possible distribution could be represented as:

∗∗∣∗∗∗∗∣∗∗∗∗∗∗∣∗∗∗∗∣∗∗∗∗

In this representation, the leftmost bin has 2 balls, the middle bin has 4 balls, and the rightmost bin has 4 balls.

Now, let's introduce the constraint that each bin can hold at most 5 balls. We can solve this by considering the maximum number of balls each bin can hold, which is 5, and then distribute the remaining balls accordingly.

Let's denote x1​,x2​,x3​ as the number of dodgeballs in bins 1, 2, and 3, respectively. We have the constraint:

x1​+x2​+x3​=13

And each xi​ must be less than or equal to 5.

Now, we can solve this using the stars and bars technique with the additional constraint. The formula for distributing identical objects into distinct bins without any constraint is (k−1n+k−1​). However, we need to adjust for the constraint. We'll subtract the cases where one or more bins contain more than 5 balls.

So, the total number of ways to distribute the dodgeballs into the bins without violating the constraint is:

Total ways=(13+3−13−1)−cases violating the constraintTotal ways=(3−113+3−1​)−cases violating the constraint

Now, let's calculate the cases where the constraint is violated:

1. If any bin contains more than 5 balls, there are no valid distributions. So, we don't need to calculate this separately.
2. If two bins each contain more than 5 balls, it's not possible. So, we only need to consider cases where one bin contains more than 5 balls.

Let's compute:

Total ways=(152)−cases where one bin contains more than 5 ballsTotal ways=(215​)−cases where one bin contains more than 5 balls

Now, we need to calculate the cases where one bin contains more than 5 balls. There are 3 bins, so there are 3 ways to choose which bin contains more than 5 balls. For each of those cases, we will distribute the balls accordingly.

Let's calculate:

Total ways=(152)−(3×ways to distribute balls where one bin contains more than 5 balls)Total ways=(215​)−(3×ways to distribute balls where one bin contains more than 5 balls)

Now, let's calculate:

Total ways=(152)−(3×ways to distribute balls where one bin contains more than 5 balls)Total ways=(215​)−(3×ways to distribute balls where one bin contains more than 5 balls) =(152)−(3×((82)+(72)+(62)))=(215​)−(3×((28​)+(27​)+(26​))) =(152)−(3×(28+21+15))=(215​)−(3×(28+21+15)) =(152)−(3×64)=(215​)−(3×64) =105−192=105−192 =−87=−87

However, the result is negative, indicating that there are no valid ways to distribute the dodgeballs into the bins without violating the constraint that each bin can hold at most 5 balls.

This means there is no solution to this problem.