Okay, so I need to figure out how many ways there are to put 5 indistinguishable balls into 2 indistinguishable boxes. Hmm, let me start by understanding the problem better. 

First, both the balls and the boxes are not distinguishable. That means if I swap the boxes, it's the same arrangement. So, the order of the boxes doesn't matter. Also, the balls are identical, so swapping balls around doesn't create a new arrangement. 

Let me recall some combinatorics principles. When arranging identical objects into identical boxes, the problem is related to integer partitions. Specifically, we need to find the number of ways to partition the integer 5 into at most 2 parts, where the order of the parts doesn't matter. 

Wait, so if we have 5 balls, we can think of distributing them into two boxes. Since the boxes are identical, the key is to find all unique pairs (a, b) where a + b = 5, and a and b are non-negative integers, but since we have two boxes, we don't have to consider a or b being zero unless allowed. Wait, the problem says "put 5 balls in 2 boxes," so each box can have zero or more balls. So, the possible distributions are all pairs (a, b) where a + b = 5, and a and b are non-negative integers. But since the boxes are identical, the pair (a, b) is the same as (b, a). Therefore, we need to count the number of unique pairs where a ≤ b to avoid duplicates.

Alternatively, since the boxes are indistinct, the number of ways is equal to the number of integer partitions of 5 into at most 2 parts. Let me recall that integer partitions of 5 are:

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

But since we have only 2 boxes, we need to consider partitions into at most 2 parts. So, the valid partitions would be:

1. 5 (which would be (5,0))
2. 4 + 1 (which is (4,1))
3. 3 + 2 (which is (3,2))

But wait, are these the only ones? Let me check. Since we can have two boxes, the maximum number of parts is 2. So, the integer partitions of 5 into exactly 1 or 2 parts. The partitions into exactly 1 part are just [5], and into exactly 2 parts are [4,1], [3,2]. So, total of 3 partitions. But wait, the problem is when boxes are indistinct, so (5,0) is considered the same as (0,5), but since we have two boxes, and we can have zero in a box, but since the boxes are identical, we need to see if (5,0) is considered different from (0,5). But since boxes are identical, no. So, (5,0) is just one unique case, where one box has all the balls and the other has none.

Similarly, (4,1) and (3,2) are two distinct cases because the partitions are different. So, total of 3 ways? Wait, but let me think again.

Alternatively, maybe the answer is 3. But let me verify with another approach. 

Another way to approach this is to model the problem as finding the number of non-negative integer solutions to the equation x + y = 5, where x and y are the number of balls in each box. Since the boxes are indistinct, each pair (x, y) is the same as (y, x). So, the number of solutions is equal to the number of distinct pairs where x ≤ y. 

So, solving x + y = 5 with x ≤ y and x, y ≥ 0. Let's list all possible pairs:

x can range from 0 up to floor(5/2) = 2.

For x=0: y=5. So, (0,5)

x=1: y=4. So, (1,4)

x=2: y=3. So, (2,3)

If x=3, y=2, but since x ≤ y, we already considered (2,3). So, total of 3 pairs. Therefore, 3 ways. 

But wait, the problem is that when boxes are identical, but if we have two boxes, can we have (0,5) and (5,0) considered the same? Yes, because swapping the boxes doesn't create a new arrangement. So, indeed, only 3 distinct ways.

But let me check another angle. Let's think of the balls as identical and the boxes as identical. So, the problem reduces to the number of distinct multisets of size 2 where the sum is 5. But since the boxes are identical, it's the same as the number of integer partitions of 5 into at most 2 parts, which is 3. So, 3 ways.

Wait, but sometimes when boxes are allowed to be empty, the formula for identical items and identical boxes is different. Let me recall the formula. 

For identical objects and identical boxes, the number of ways is equal to the number of integer partitions of n into at most k parts, where k is the number of boxes. But in this case, n=5 and k=2. So, the number of partitions of 5 into at most 2 parts. Which are:

1. 5

2. 4 + 1

3. 3 + 2

So, 3 partitions. Therefore, the answer is 3.

But wait, let me confirm with a concrete example. Let's list all possible distributions:

1. All 5 balls in one box, 0 in the other. Since boxes are identical, this is one case.

2. 4 balls in one box, 1 in the other. Since the boxes are identical, (4,1) is the same as (1,4), so this is another case.

3. 3 balls in one box, 2 in the other. Again, since boxes are identical, this is a third case.

What about other distributions, like 3,1,1? Wait, but since we only have two boxes, we can't have three balls in one box and then two balls split into two boxes. Wait, no, the problem is to distribute into two boxes. So, if we have 3 and 2, that's the maximum we can do with two boxes. So, the possible partitions are indeed 5; 4+1; 3+2. 

Therefore, the answer should be 3.

But let me think again. Suppose we have two boxes, and the balls are identical. The possible distributions are:

- Both boxes have 0 and 5 balls.

- One box has 1, the other has 4.

- One box has 2, the other has 3.

But wait, can we have a distribution like 3,1,1? No, because we only have two boxes. So, the third ball would have to go into another box, but we only have two. So, the maximum number of balls in a single box is 5, then 4, then 3. 

Therefore, the possible distributions are the ones I listed earlier: (5,0), (4,1), (3,2). So, 3 ways.

Alternatively, if we consider that boxes are distinguishable, the number would be 6, since for each distribution (a,b), with a + b =5, there are two possibilities unless a = b. But since the boxes are indistinct, we have to count each pair only once. So, the formula for indistinct boxes and identical objects is the number of integer partitions of 5 into at most 2 parts, which is 3.

Therefore, the answer is 3.

But to be thorough, let's list all possible distributions:

1. [5,0]

2. [4,1]

3. [3,2]

These are the three distinct ways, considering that the boxes are identical. If the boxes were distinct, there would be 6 ways (since each distribution is counted twice except when a = b, but here 5 and 0, 4 and 1, 3 and 2 are all different, so 2*3=6). But since the boxes are identical, we have 3 ways.

Therefore, the answer is 3. So, the number of ways is \boxed{3}.
Therefore, the number of ways to distribute 5 indistinguishable balls into 2 indistinguishable boxes is $\boxed{3}$.
</think>