Distributing Balls into Bags: A Comprehensive Approach to Solving the Problem with the Principle of Inclusion-Exclusion

In this article, we will delve into a fascinating mathematical problem: how many ways can 7 balls, where 3 are identical and 4 are distinct, be distributed into 7 bags such that no bag is left empty? We'll explore step-by-step the application of the principle of inclusion-exclusion to solve this problem and ensure the content aligns with Google's search engine optimization (SEO) standards.

Step 1: Total Distributions Without Restrictions

Let's start by calculating the total number of ways to distribute 7 balls into 7 bags without any restrictions. Since each ball can go into any of the 7 bags, the total number of distributions is:

77

Step 2: Exclude Distributions with Empty Bags

Next, we need to exclude the distributions where at least one bag is empty. This is where the principle of inclusion-exclusion comes into play.

Let S represent the set of all distributions of 7 balls into 7 bags. We denote S as the total number of ways to distribute the balls without restrictions, which is 77.

Let Ai be the set of distributions where bag i is empty. We aim to find S - A1 ∪ A2 ∪ ... ∪ A7.

Using the principle of inclusion-exclusion:

A1 ∪ A2 ∪ ... ∪ A7 Σi17 Ai - Σ1 ≤j 7 Ai ∩ Aj Σ1 ≤j k 7 Ai ∩ Aj ∩ Ak - ... - (?1)7 A1 ∩ A2 ∩ ... ∩ A7

Step 3: Calculate Each Term

Single Empty Bag:

When one bag is empty, we distribute 7 balls into 6 bags.

Ai 67

There are 7 such bags, so:

Σi17 Ai 7 × 67

Two Empty Bags:

If two bags are empty, we distribute 7 balls into 5 bags.

Ai ∩ Aj 57

There are (binom{7}{2}) ways to choose 2 bags to be empty, so:

Σ1 ≤j 7 Ai ∩ Aj (binom{7}{2}) × 57 21 × 57

Three Empty Bags:

If three bags are empty, we distribute 7 balls into 4 bags.

Ai ∩ Aj ∩ Ak 47

There are (binom{7}{3}) ways to choose 3 bags to be empty, so:

Σ1 ≤j k 7 Ai ∩ Aj ∩ Ak (binom{7}{3}) × 47 35 × 47

Four Empty Bags:

With four bags empty, we distribute 7 balls into 3 bags.

Ai ∩ Aj ∩ Ak ∩ Al 37

There are (binom{7}{4}) ways to choose 4 bags to be empty, so:

Σ1 ≤j k l 7 Ai ∩ Aj ∩ Ak ∩ Al (binom{7}{4}) × 37 35 × 37

Five Empty Bags:

For five bags empty, we distribute 7 balls into 2 bags.

Ai ∩ Aj ∩ Ak ∩ Al ∩ Am 27

There are (binom{7}{5}) ways to choose 5 bags to be empty, so:

Σ1 ≤j k l m 7 Ai ∩ Aj ∩ Ak ∩ Al ∩ Am (binom{7}{5}) × 27 21 × 27

Six Empty Bags:

When six bags are empty, we distribute 7 balls into 1 bag.

Ai ∩ Aj ∩ Ak ∩ Al ∩ Am ∩ An 17 1

There are (binom{7}{6}) ways to choose 6 bags to be empty, so:

Σ1 ≤j k l m n 7 Ai ∩ Aj ∩ Ak ∩ Al ∩ Am ∩ An (binom{7}{6}) × 17 7

Step 4: Putting It All Together

Now we can substitute everything into our inclusion-exclusion formula:

A1 ∪ A2 ∪ ... ∪ A7 7 × 67 - 21 × 57 35 × 47 - 35 × 37 21 × 27 - 7

Finally, the number of ways to distribute the balls such that no bag is empty is:

S - A1 ∪ A2 ∪ ... ∪ A7 77 - 7 × 67 - 21 × 57 35 × 47 - 35 × 37 21 × 27 - 7

Conclusion

By computing this final expression, you can determine the total number of ways to distribute the 7 balls into 7 bags with no bag left empty.