How many ways are there to pack six copies of the same book into four identical boxes, where a box can contain as many as six books ?
9
Understanding the Problem: Packing Identical Books
This question asks about the number of ways to distribute identical items (six copies of the same book) into identical containers (four identical boxes). The key words here are "identical books" and "identical boxes". This type of problem is a classic example of integer partitioning in combinatorics.
Since the books are identical, the order in which we place them in a box doesn't matter, and which specific book goes into which box doesn't matter. Since the boxes are identical, swapping the contents of two boxes does not result in a new way of packing if the set of contents in each box is the same. For example, packing (3, 3, 0, 0) is the same as (0, 3, 3, 0) because the boxes are identical.
A box can contain up to six books. Since there are only six books in total, any box can technically hold up to all six books, so this constraint is not limiting in this specific case. The total number of books packed must be six.
Relating to Integer Partitions
The problem can be framed as finding the number of ways to partition the integer 6 into at most 4 parts. Each part in the partition represents the number of books in a non-empty box. Since we have 4 boxes, we can use 1, 2, 3, or 4 non-empty boxes. The sum of the numbers of books in the non-empty boxes must equal 6.
Let's list the partitions of the integer 6 and see how many parts each partition has:
| Partition of 6 | Number of Parts | Interpretation (Books per box in non-empty boxes) | Valid for 4 Boxes? (Parts $\le$ 4) |
|---|---|---|---|
| 6 | 1 | (6) - One box with 6 books | Yes |
| 5 + 1 | 2 | (5, 1) - One box with 5, one with 1 | Yes |
| 4 + 2 | 2 | (4, 2) - One box with 4, one with 2 | Yes |
| 3 + 3 | 2 | (3, 3) - Two boxes with 3 books each | Yes |
| 4 + 1 + 1 | 3 | (4, 1, 1) - One box with 4, two with 1 | Yes |
| 3 + 2 + 1 | 3 | (3, 2, 1) - One box with 3, one with 2, one with 1 | Yes |
| 2 + 2 + 2 | 3 | (2, 2, 2) - Three boxes with 2 books each | Yes |
| 3 + 1 + 1 + 1 | 4 | (3, 1, 1, 1) - One box with 3, three with 1 | Yes |
| 2 + 2 + 1 + 1 | 4 | (2, 2, 1, 1) - Two boxes with 2, two with 1 | Yes |
| 2 + 1 + 1 + 1 + 1 | 5 | (2, 1, 1, 1, 1) - One box with 2, four with 1 | No (Requires 5 boxes) |
| 1 + 1 + 1 + 1 + 1 + 1 | 6 | (1, 1, 1, 1, 1, 1) - Six boxes with 1 book each | No (Requires 6 boxes) |
Counting the Valid Partitions
We are looking for partitions of 6 that have at most 4 parts. From the table above, these are the partitions with 1, 2, 3, or 4 parts:
- Partitions with 1 part: 6 (1 way)
- Partitions with 2 parts: 5+1, 4+2, 3+3 (3 ways)
- Partitions with 3 parts: 4+1+1, 3+2+1, 2+2+2 (3 ways)
- Partitions with 4 parts: 3+1+1+1, 2+2+1+1 (2 ways)
The total number of ways to pack the six identical books into four identical boxes is the sum of the number of ways for each possible number of non-empty boxes (from 1 to 4).
Total ways = (Ways with 1 part) + (Ways with 2 parts) + (Ways with 3 parts) + (Ways with 4 parts)
Total ways = $1 + 3 + 3 + 2 = 9$
Conclusion
There are 9 distinct ways to pack six identical copies of the same book into four identical boxes, allowing for empty boxes.
Revision Table: Packing Identical Items
| Concept | Description | Application Here |
|---|---|---|
| Integer Partition | A way of writing a positive integer as a sum of positive integers. The order of the summands (parts) does not matter. | We partition the total number of books (6). |
| Identical Items | Items that are indistinguishable from one another. | The six copies of the book are identical. |
| Identical Containers | Containers that are indistinguishable from one another. | The four boxes are identical. |
| Packing Problem | Distributing items into containers. | Distributing 6 identical books into 4 identical boxes. |
| Parts of a Partition | The positive integers that sum up to the original integer. Corresponds to the number of items in each non-empty container when items and containers are identical. | The number of books in each non-empty box. |
| Partitions into $\le k$ parts | Partitions where the number of summands is $k$ or less. | The number of non-empty boxes used is 4 or less. |
Additional Information: Partition Function
The number of partitions of an integer $n$ is denoted by $p(n)$. There is no simple closed-form formula for $p(n)$, but it can be calculated recursively or using generating functions. The question here is about $p(n, \le k)$, the number of partitions of $n$ into at most $k$ parts. A useful identity is that $p(n, \le k)$ is equal to the number of partitions of $n$ where the largest part is at most $k$. For this problem ($n=6, k=4$), this means the number of partitions of 6 where the largest part is $\le 4$. Let's check this:
- Partitions of 6: 6, 5+1, 4+2, 3+3, 4+1+1, 3+2+1, 2+2+2, 3+1+1+1, 2+2+1+1, 2+1+1+1+1, 1+1+1+1+1+1.
- Partitions where the largest part is $\le 4$:
- 4+2 (Largest part 4)
- 3+3 (Largest part 3)
- 4+1+1 (Largest part 4)
- 3+2+1 (Largest part 3)
- 2+2+2 (Largest part 2)
- 3+1+1+1 (Largest part 3)
- 2+2+1+1 (Largest part 2)
- 2+1+1+1+1 (Largest part 2)
- 1+1+1+1+1+1 (Largest part 1)
Counting these partitions: There are 9 such partitions. This confirms the result obtained by counting partitions into at most 4 parts.
Important Questions from Combinations
Let us assume a person climbing the stairs can take one stair or two stairs at a time. How many ways can this person climb a flight of eight stairs?
There are 9 different rock lizard species, each with a unique colour. Lizards of 2 different species sit on a rock at any given time. The number of possible colour combinations of rock lizards seen together on a rock is _________
(Answer in integer)- There are nine species of Impatiens (balsams) found in laterite plateaus of the northern Western Ghats, each with a distinct colour. If a plateau has exactly 6 species, then the number of possible colour combinations in the plateau is __________. (Answer in integer)
- The number of different possible ways of forming five intramolecular disulfide bonds with ten cysteine residues of a protein is ________
- A set of 4 parallel lines intersect with another set of 5 parallel lines. How many parallelograms are formed?
