Distributing 10 Distinguishable Books Into 5 Indistinguishable Boxes A Combinatorial Challenge

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Hey guys! 👋 Today, we're diving into a super interesting math problem that combines combinatorics and a bit of strategic thinking. We're going to figure out how many ways we can distribute 10 unique books into 5 identical boxes, making sure each box has at least one book. Buckle up, because this is going to be a fun ride!

Breaking Down the Problem

So, let's get our heads around the question: How many ways can we put 10 distinguishable books into 5 indistinguishable boxes so that each box contains at least one book?

Before we jump into calculations, let's make sure we understand the key elements. We have 10 different books – think of them as each having a unique title and cover. Then, we have 5 boxes, but here's the twist: they're identical. This means swapping the contents of two boxes doesn't create a new arrangement. Finally, each box needs to have at least one book. This last condition is crucial because it changes how we approach the problem.

To really nail this, we need to consider the different ways we can split the 10 books into 5 groups. We need to figure out which combinations of group sizes are possible. For example, we could have groups of 6, 1, 1, 1, and 1 books. Or maybe 5, 2, 1, 1, and 1. You get the idea! Once we know the possible group sizes, we can calculate the number of ways to create those groups and then account for the fact that the boxes are indistinguishable.

Understanding the Constraints

The core challenge here is ensuring every box has at least one book. If we didn't have this restriction, the problem would be a classic "stars and bars" scenario, but with a twist due to the indistinguishable boxes. However, the "at least one book per box" rule means we need a more nuanced approach. This constraint significantly reduces the number of possible distributions because it eliminates scenarios where one or more boxes are empty.

We also need to remember that the books are distinguishable, meaning that the order of books within a box doesn't matter, but the specific books in each box do matter. If we put books A, B, and C in one box, it's different from putting books D, E, and F in that same box. This is why we'll be using combinations (choosing a subset of books) rather than permutations (arranging books in a specific order).

The fact that the boxes are indistinguishable adds another layer of complexity. If the boxes were distinguishable (say, they were labeled Box 1, Box 2, etc.), we would simply multiply the number of ways to form the groups by the number of ways to arrange the groups among the boxes (which would be 5! in some cases). But because the boxes are identical, we need to avoid overcounting arrangements that are just rearrangements of the same groups. This is a key point to keep in mind as we work through the solution.

Possible Distributions

First, we need to figure out the possible ways to divide 10 books into 5 non-empty groups. Think of it like this: what are the different combinations of numbers that add up to 10, where we have exactly 5 numbers, and each number is at least 1? Let's list them out:

  1. 6 + 1 + 1 + 1 + 1
  2. 5 + 2 + 1 + 1 + 1
  3. 4 + 3 + 1 + 1 + 1
  4. 4 + 2 + 2 + 1 + 1
  5. 3 + 3 + 2 + 1 + 1
  6. 3 + 2 + 2 + 2 + 1
  7. 2 + 2 + 2 + 2 + 2

These are all the possible ways to partition the 10 books into 5 groups, ensuring each group has at least one book. Now, for each of these partitions, we need to calculate how many ways we can actually form those groups from our 10 distinct books.

Calculating the Combinations

Now, let's calculate the number of ways to arrange the books for each of these distributions. This is where the combinations come in – we'll be using the formula for combinations, which is:

C(n, k) = n! / (k! * (n - k)!)

Where:

  • n is the total number of items
  • k is the number of items to choose
  • ! denotes factorial (e.g., 5! = 5 × 4 × 3 × 2 × 1)

Let's go through each case:

Case 1: 6 + 1 + 1 + 1 + 1

  • Ways to choose 6 books: C(10, 6) = 10! / (6! * 4!) = 210
  • Ways to choose 1 book from remaining 4: C(4, 1) = 4
  • Ways to choose 1 book from remaining 3: C(3, 1) = 3
  • Ways to choose 1 book from remaining 2: C(2, 1) = 2
  • Ways to choose 1 book from remaining 1: C(1, 1) = 1

Multiplying these together: 210 * 4 * 3 * 2 * 1 = 5040. But since the four groups of 1 are indistinguishable, we divide by 4! (the number of ways to arrange the four groups of 1): 5040 / 4! = 5040 / 24 = 210.

So, there are 210 ways for this case.

Case 2: 5 + 2 + 1 + 1 + 1

  • Ways to choose 5 books: C(10, 5) = 10! / (5! * 5!) = 252
  • Ways to choose 2 books from remaining 5: C(5, 2) = 10
  • Ways to choose 1 book from remaining 3: C(3, 1) = 3
  • Ways to choose 1 book from remaining 2: C(2, 1) = 2
  • Ways to choose 1 book from remaining 1: C(1, 1) = 1

Multiplying these together: 252 * 10 * 3 * 2 * 1 = 15120. Divide by 3! (for the three groups of 1): 15120 / 6 = 2520

So, there are 2520 ways for this case.

Case 3: 4 + 3 + 1 + 1 + 1

  • Ways to choose 4 books: C(10, 4) = 210
  • Ways to choose 3 books from remaining 6: C(6, 3) = 20
  • Ways to choose 1 book from remaining 3: C(3, 1) = 3
  • Ways to choose 1 book from remaining 2: C(2, 1) = 2
  • Ways to choose 1 book from remaining 1: C(1, 1) = 1

Multiply: 210 * 20 * 3 * 2 * 1 = 25200. Divide by 3! (for the three groups of 1): 25200 / 6 = 4200

4200 ways for this case.

Case 4: 4 + 2 + 2 + 1 + 1

  • Ways to choose 4 books: C(10, 4) = 210
  • Ways to choose 2 books from remaining 6: C(6, 2) = 15
  • Ways to choose 2 books from remaining 4: C(4, 2) = 6
  • Ways to choose 1 book from remaining 2: C(2, 1) = 2
  • Ways to choose 1 book from remaining 1: C(1, 1) = 1

Multiply: 210 * 15 * 6 * 2 * 1 = 37800. Divide by 2! for the two groups of 2 and 2! for the two groups of 1: 37800 / (2 * 2) = 9450

9450 ways for this case.

Case 5: 3 + 3 + 2 + 1 + 1

  • Ways to choose 3 books: C(10, 3) = 120
  • Ways to choose 3 books from remaining 7: C(7, 3) = 35
  • Ways to choose 2 books from remaining 4: C(4, 2) = 6
  • Ways to choose 1 book from remaining 2: C(2, 1) = 2
  • Ways to choose 1 book from remaining 1: C(1, 1) = 1

Multiply: 120 * 35 * 6 * 2 * 1 = 50400. Divide by 2! for the two groups of 3 and 2! for the two groups of 1: 50400 / (2 * 2) = 12600

12600 ways for this case.

Case 6: 3 + 2 + 2 + 2 + 1

  • Ways to choose 3 books: C(10, 3) = 120
  • Ways to choose 2 books from remaining 7: C(7, 2) = 21
  • Ways to choose 2 books from remaining 5: C(5, 2) = 10
  • Ways to choose 2 books from remaining 3: C(3, 2) = 3
  • Ways to choose 1 book from remaining 1: C(1, 1) = 1

Multiply: 120 * 21 * 10 * 3 * 1 = 75600. Divide by 3! for the three groups of 2: 75600 / 6 = 12600

12600 ways for this case.

Case 7: 2 + 2 + 2 + 2 + 2

  • Ways to choose 2 books: C(10, 2) = 45
  • Ways to choose 2 books from remaining 8: C(8, 2) = 28
  • Ways to choose 2 books from remaining 6: C(6, 2) = 15
  • Ways to choose 2 books from remaining 4: C(4, 2) = 6
  • Ways to choose 2 books from remaining 2: C(2, 2) = 1

Multiply: 45 * 28 * 15 * 6 * 1 = 113400. Divide by 5! for the five groups of 2: 113400 / 120 = 945

945 ways for this case.

Summing It Up

Now, we add up the results from all the cases:

210 + 2520 + 4200 + 9450 + 12600 + 12600 + 945 = 42525

Final Answer

So, there are a grand total of 42,525 ways to put 10 distinguishable books into 5 indistinguishable boxes, with each box containing at least one book.

Isn't that a cool problem? It really shows how combinatorics can get intricate, and how important it is to break down a problem into smaller, manageable steps. We had to consider the constraints, figure out the possible distributions, calculate combinations, and then avoid overcounting. It's like a puzzle with many layers!

Hope you guys enjoyed this journey into the world of combinatorics! Keep exploring, keep questioning, and most importantly, keep having fun with math! 🎉