In how many ways can 20 identical coins be distributed into 6 different bags such that no bag contains more than 5 coins?

We are given the problem of distributing 20 identical coins into 6 different bags such that no bag contains more than 5 coins. The goal is to find the total number of such distributions.

### Step 1: Problem Representation
Let \( x_1, x_2, \dots, x_6 \) be the number of coins in the 6 bags. The problem then translates to finding the number of non-negative integer solutions to the equation:

\[
x_1 + x_2 + \dots + x_6 = 20
\]

subject to the constraints:

\[
0 \leq x_i \leq 5 \quad \text{for all } i = 1, 2, \dots, 6.
\]

### Step 2: Generating Function Approach
One effective method to solve this problem is to use generating functions. The generating function for the number of coins in each bag \( x_i \) can be expressed as:

\[
G(x) = (1 + x + x^2 + x^3 + x^4 + x^5)
\]

We seek the coefficient of \( x^{20} \) in the expansion of \( G(x)^6 \). Let's express \( G(x) \) as a closed-form:

\[
G(x) = \frac{1-x^6}{1-x}
\]

Thus, the generating function for all 6 bags is:

\[
\left( \frac{1-x^6}{1-x} \right)^6 = \frac{(1-x^6)^6}{(1-x)^6}
\]

### Step 3: Expand and Extract Coefficient
We need to expand this expression and find the coefficient of \( x^{20} \).

The expansion of \( (1-x^6)^6 \) can be done using the binomial theorem:

\[
(1 - x^6)^6 = \sum_{k=0}^{6} \binom{6}{k} (-1)^k x^{6k}
\]

The expansion of \( (1-x)^{-6} \) is a known series:

\[
(1-x)^{-6} = \sum_{m=0}^{\infty} \binom{m+5}{5} x^m
\]

Now, the generating function we need is:

\[
\sum_{k=0}^{6} \binom{6}{k} (-1)^k x^{6k} \times \sum_{m=0}^{\infty} \binom{m+5}{5} x^m
\]

We need to find the coefficient of \( x^{20} \) in this product, which is:

\[
\sum_{k=0}^{6} \binom{6}{k} (-1)^k \binom{25-6k}{5}
\]

### Step 4: Compute the Required Coefficient

Let's calculate this sum:

\[
\text{Coefficient of } x^{20} = \sum_{k=0}^{6} \binom{6}{k} (-1)^k \binom{25-6k}{5}
\]

Evaluating each term:

- For \( k = 0 \), \( \binom{25}{5} = 53130 \)
- For \( k = 1 \), \( \binom{19}{5} = 11628 \)
- For \( k = 2 \), \( \binom{13}{5} = 1287 \)
- For \( k = 3 \), \( \binom{7}{5} = 21 \)
- For \( k = 4 \), \( \binom{1}{5} = 0 \) (since this is undefined, treat as 0)
- For \( k = 5 \), \( \binom{-5}{5} = 0 \) (since this is undefined, treat as 0)
- For \( k = 6 \), \( \binom{-11}{5} = 0 \) (since this is undefined, treat as 0)

Thus, the sum is:

\[
\text{Coefficient of } x^{20} = 53130 - 11628 + 1287 - 21 + 0 + 0 + 0 = 42768
\]

### Final Answer:
The number of ways to distribute 20 identical coins into 6 different bags such that no bag contains more than 5 coins is **42,768**.

Would you like further details or have any questions?

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Here are 8 related questions:

1. How would the solution change if there were 5 bags instead of 6?
2. What if the maximum number of coins per bag were increased to 6?
3. How many ways are there if exactly one bag must contain exactly 5 coins?
4. What if the problem were to distribute 25 coins instead of 20?
5. How would the generating function approach change if the bags were indistinguishable?
6. Can the problem be solved using a recursive approach instead of generating functions?
7. How does the problem relate to combinatorial partitions?
8. How would the solution change if some of the bags were pre-filled with a certain number of coins?

**Tip:** Always ensure that you correctly handle constraints in combinatorial problems, as overlooking them can lead to incorrect solutions.

Explore the mathematical problem of distributing 20 identical coins into 6 different bags, ensuring no bag contains more than 5 coins. Learn the application of generating functions and binomial coefficients in solving combinatorial problems. Suitable for advanced high school students.

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