This problem involves distributing 4 indistinguishable balls into 4 distinguishable boxes. The key point here is that the balls are indistinguishable, meaning we're only concerned with how many balls go into each box, not the order of the balls themselves.

This type of problem is often referred to as the stars and bars problem, which is a combinatorics method used to determine how to distribute indistinguishable objects (stars) into distinct groups (bars).

The formula for stars and bars is:

$\text{Number of ways} = \binom{n + k - 1}{k - 1}$

Where:

• $n$ is the number of indistinguishable objects (balls),
• $k$ is the number of distinguishable groups (boxes),
• $\binom{n + k - 1}{k - 1}$ is the binomial coefficient, or "n choose k."

In this case, $n = 4$ (balls) and $k = 4$ (boxes). So, the number of ways is:

$\binom{4 + 4 - 1}{4 - 1} = \binom{7}{3}$

Now, calculate $\binom{7}{3}$:

$\binom{7}{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$

So, the number of ways to place 4 indistinguishable balls into 4 distinguishable boxes is 35.

Would you like further clarification or details on this solution?

Here are some related questions you might find helpful:

1. How does the stars and bars method change if the boxes are also indistinguishable?
2. What if each box must contain at least one ball? How many ways are there then?
3. Can the stars and bars method be applied if the balls are distinguishable?
4. How many ways can you distribute 6 balls into 3 distinguishable boxes?
5. What if we allow some boxes to be empty, does the method still work?

Tip: The stars and bars formula is a powerful tool for solving partitioning problems in combinatorics—practice identifying when you can use it!