To determine the number of ways to distribute 20 identical coins into 6 different bags such that no bag contains more than 5 coins, we can use the method of generating functions.

Step 1: Problem Representation

Let $x_1, x_2, \dots, x_6$ be the number of coins in the 6 bags. The problem then is to find the number of non-negative integer solutions to the equation:

$x_1 + x_2 + \dots + x_6 = 20$

with the constraint:

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

Step 2: Generating Function Approach

The generating function for each $x_i$ is:

$G(x) = 1 + x + x^2 + x^3 + x^4 + x^5 = \frac{1-x^6}{1-x}$

We want to find the coefficient of $x^{20}$ in the expansion of $G(x)^6$, which is:

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

Step 3: Expand and Extract Coefficient

Expand $(1-x^6)^6$ 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:

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

Now, multiply these two expansions to find the coefficient of $x^{20}$:

$\text{Coefficient of } x^{20} = \sum_{k=0}^{3} \binom{6}{k} (-1)^k \binom{25-6k}{5}$

Step 4: Calculate the Required Coefficient

Let's calculate this sum:

• For $k = 0$: $\binom{6}{0} \times \binom{25}{5} = 1 \times 53130 = 53130$
• For $k = 1$: $\binom{6}{1} \times \binom{19}{5} = 6 \times 11628 = 69768$
• For $k = 2$: $\binom{6}{2} \times \binom{13}{5} = 15 \times 1287 = 19305$
• For $k = 3$: $\binom{6}{3} \times \binom{7}{5} = 20 \times 21 = 420$

Substituting into the sum:

$53130 - 69768 + 19305 - 420 = 32247$

Thus, the number of ways to distribute the coins is 32,247.

Would you like more 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 does the generating function approach change if the bags were indistinguishable?
6. Can this problem be solved using dynamic programming?
7. How does the solution relate to combinatorial partitions?
8. How would the solution change if each bag had to contain at least 1 coin?

Tip: Double-check your calculations in combinatorial problems, especially when subtracting cases that don’t satisfy the constraints.