A bag contains 3 black, 4 white and 2 red balls, all the balls being different. The number of at most 6 balls containing balls of all the colours is

Options

A. `42 (4!)`
B. ` 2^6 xx 4!`
C. ` (2^6 -1)(4!)`
D. None of these

Answer

To solve the problem, we need to find the number of ways to select at most 6 balls from a bag containing 3 black, 4 white, and 2 red balls, ensuring that we select at least one ball of each color.

### Step-by-Step Solution:

1. **Identify the total number of balls**:
- Black balls = 3
- White balls = 4
- Red balls = 2
- Total = 3 + 4 + 2 = 9 balls.

2. **Understand the selection criteria**:
- We need to select at most 6 balls.
- We must include at least one ball of each color (black, white, red).

3. **Define the selection cases**:
- Since we need at least one ball of each color, we can denote the number of balls selected from each color as follows:
- Let \( b \) = number of black balls selected
- Let \( w \) = number of white balls selected
- Let \( r \) = number of red balls selected
- The conditions are:
- \( b + w + r \leq 6 \)
- \( b \geq 1 \), \( w \geq 1 \), \( r \geq 1 \)

4. **Transform the selection variables**:
- To simplify, let’s redefine the variables:
- Let \( b' = b - 1 \) (so \( b' \geq 0 \))
- Let \( w' = w - 1 \) (so \( w' \geq 0 \))
- Let \( r' = r - 1 \) (so \( r' \geq 0 \))
- Now, we can rewrite the selection condition as:
- \( (b' + 1) + (w' + 1) + (r' + 1) \leq 6 \)
- This simplifies to \( b' + w' + r' \leq 3 \).

5. **Count the valid selections**:
- Now we need to count the non-negative integer solutions to the equation \( b' + w' + r' \leq 3 \).
- We can introduce a new variable \( x \) such that \( b' + w' + r' + x = 3 \), where \( x \) represents the unused selections.
- The total number of non-negative integer solutions to this equation is given by the stars and bars theorem:
- The number of solutions is \( \binom{3 + 4 - 1}{4 - 1} = \binom{6}{3} = 20 \).

6. **Calculate the combinations for each case**:
- For each valid combination of \( b, w, r \) (where \( b \) can be from 1 to 3, \( w \) from 1 to 4, and \( r \) from 1 to 2), we need to calculate the number of ways to choose the balls:
- For each combination of \( b, w, r \), the number of ways to choose the balls is given by:
- \( \binom{3}{b} \times \binom{4}{w} \times \binom{2}{r} \).

7. **Sum all valid combinations**:
- We need to sum the products of the combinations for all valid \( b, w, r \) selections that satisfy the conditions.

8. **Final count**:
- After calculating all valid combinations and their respective counts, we arrive at the final answer.