Number of ways in which 15 identical coins can be put into 6 different bags

Options

A. is coefficient of `x^(15)` in `x^(6)(1+x+x^(2)+ . . .oo)^(6),` if no bag remains empty
B. is coefficient of `x^(15)` in `(1-x)^(-6)`
C. is same as number of the integral solutions of
`a+b+c+d+e+f=15`
D. is same as number of non-negative integral solutions of `underset(i=1)overset(6)(sum)x_(i)=15`

Explanation

To solve the problem of distributing 15 identical coins into 6 different bags, we can use the "stars and bars" theorem from combinatorics.

### Step-by-Step Solution:

1. **Understanding the Problem**: We need to distribute 15 identical coins into 6 different bags. This can be represented mathematically as finding the number of non-negative integer solutions to the equation:
\[ x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 15 \]
where \(x_i\) represents the number of coins in bag \(i\).

2. **Applying the Stars and Bars Theorem**: The stars and bars theorem states that the number of ways to put \(n\) indistinguishable objects (stars) into \(k\) distinguishable boxes (bars) is given by:
\[ \binom{n + k - 1}{k - 1} \]
In our case, \(n = 15\) (the coins) and \(k = 6\) (the bags).

3. **Calculating the Values**: Plugging in the values into the formula:
\[ \binom{15 + 6 - 1}{6 - 1} = \binom{20}{5} \]

4. **Calculating the Binomial Coefficient**:
\[ \binom{20}{5} = \frac{20!}{5!(20-5)!} = \frac{20!}{5! \cdot 15!} \]
This can be simplified as follows:
\[ \binom{20}{5} = \frac{20 \times 19 \times 18 \times 17 \times 16}{5 \times 4 \times 3 \times 2 \times 1} \]

5. **Performing the Calculation**:
- Calculate the numerator:
\[ 20 \times 19 = 380 \]
\[ 380 \times 18 = 6840 \]
\[ 6840 \times 17 = 116280 \]
\[ 116280 \times 16 = 1860480 \]
- Calculate the denominator:
\[ 5 \times 4 \times 3 \times 2 \times 1 = 120 \]
- Now divide the numerator by the denominator:
\[ \frac{1860480}{120} = 15504 \]

6. **Final Answer**: Therefore, the number of ways to distribute 15 identical coins into 6 different bags is:
\[ \boxed{15504} \]