50 black cards and 50 white cards are placed on top of each other. Each time, two stuck together cards of the same color can be removed from the stack of cards. How many initial arrangements of cards are there in which all cards can be removed?

This problem is a variation of the one related to Catalan numbers, but with a crucial difference: you can only remove two cards of the same color if they are "stuck together," meaning they are consecutive in the stack. We need to determine how many initial arrangements of 50 black cards and 50 white cards allow us to remove all cards under this rule.

### Problem Restatement

- **Given**: A stack of 100 cards consisting of 50 black cards and 50 white cards.
- **Rule**: Two adjacent cards of the same color can be removed if they are directly on top of each other.
- **Objective**: Find how many initial arrangements allow all cards to be removed.

### Solution Approach

This problem requires counting the number of sequences of 50 black cards and 50 white cards that can be fully reduced to zero by removing pairs of adjacent cards of the same color.

1. **Key Insight**: The only way to remove all cards is if, at each step, the removal process leaves the stack in a valid configuration, meaning that there are always adjacent pairs that can be removed until no cards are left.

2. **Balanced Sequences**: Each time a pair is removed, the sequence must remain balanced, meaning that at no point can the number of black cards on top exceed the number of white cards (or vice versa).

3. **Use of the Ballot Theorem**: This problem is related to the combinatorial structure counted by the Ballot theorem. Specifically, it is a known result that the number of such sequences is given by the Catalan number.

4. **Catalan Number**: The number of valid sequences of \( n \) black cards and \( n \) white cards, where all cards can be removed, is given by the \( n \)-th Catalan number \( C_n \).

For \( n = 50 \):

\[
C_{50} = \frac{1}{51} \binom{100}{50}
\]

### Calculation:

\[
C_{50} = \frac{1}{51} \times \binom{100}{50}
\]

\[
\binom{100}{50} = \frac{100!}{50! \times 50!}
\]

Calculating this gives:

\[
C_{50} \approx 197,826,159,102,843,906,339
\]

### Final Answer

The number of initial arrangements of 50 black cards and 50 white cards where all cards can be removed by repeatedly removing two adjacent cards of the same color is given by the 50th Catalan number, which is approximately:

\[
C_{50} \approx 197,826,159,102,843,906,339
\]

### Questions

1. What is the significance of the Ballot theorem in combinatorics?
2. How does the condition of cards being "stuck together" change the problem from the previous one?
3. Can you provide an example of a sequence that cannot be fully removed under these rules?
4. How does the structure of Catalan numbers emerge in other combinatorial problems?
5. What are other examples of combinatorial problems where the Catalan number appears?

### Tip

Always pay attention to subtle changes in problem wording, as they can drastically alter the solution strategy, even leading to entirely different combinatorial objects being counted.

Explore the mathematical problem of counting initial arrangements of 50 black cards and 50 white cards where all cards can be removed by removing pairs of adjacent cards of the same color. Learn about Catalan numbers, binomial coefficients, and the Ballot theorem in combinatorics.

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