In how many ways can five balls be taken out from a bag consisting of six identical red balls, seven identical white balls and three identical green balls?

Answer

To solve the problem of how many ways five balls can be taken out from a bag containing six identical red balls, seven identical white balls, and three identical green balls, we can follow these steps:

### Step 1: Define the Variables
Let:
- \( r \) = number of red balls taken
- \( w \) = number of white balls taken
- \( g \) = number of green balls taken

We need to satisfy the equation:
\[ r + w + g = 5 \]

### Step 2: Set Constraints
From the problem, we know:
- \( 0 \leq r \leq 6 \) (since there are 6 red balls)
- \( 0 \leq w \leq 7 \) (since there are 7 white balls)
- \( 0 \leq g \leq 3 \) (since there are 3 green balls)

### Step 3: Use the Stars and Bars Method
The problem can be approached using the "stars and bars" theorem, which is a common combinatorial method for distributing indistinguishable objects (balls) into distinguishable boxes (colors).

The formula for the number of non-negative integer solutions to the equation \( x_1 + x_2 + ... + x_k = n \) is given by:
\[ \binom{n + k - 1}{k - 1} \]
where \( n \) is the total number of items to distribute, and \( k \) is the number of categories.

In our case:
- \( n = 5 \) (the total number of balls)
- \( k = 3 \) (the three colors: red, white, green)

Thus, the number of unrestricted solutions is:
\[ \binom{5 + 3 - 1}{3 - 1} = \binom{7}{2} \]

### Step 4: Calculate the Combinations
Calculating \( \binom{7}{2} \):
\[ \binom{7}{2} = \frac{7 \times 6}{2 \times 1} = 21 \]

### Step 5: Exclude Invalid Cases
Now we need to exclude cases where the number of balls exceeds the available quantity:
1. **Case of 5 Green Balls**: Not possible since we only have 3 green balls.
2. **Case of 4 Green Balls**: Not possible since we only have 3 green balls.
3. **Case of 3 Green Balls**: If we take 3 green balls, then we can take 2 more balls from red or white. This gives us combinations of (2 red, 0 white) and (1 red, 1 white) and (0 red, 2 white). This is valid.
4. **Case of 4 Red Balls**: If we take 4 red balls, we can only take 1 more from white.
5. **Case of 4 White Balls**: If we take 4 white balls, we can only take 1 more from red.

Thus, we need to subtract the invalid cases:
- The only invalid case is taking 5 green balls, which is not possible.

### Step 6: Final Calculation
So we have:
\[ \text{Valid combinations} = 21 - 1 = 20 \]

### Conclusion
The total number of ways to take out 5 balls from the bag is **20**.

---