A box contains 4 white balls, 5 black balls and 2 red balls. The number of ways three balls be drawn from the box, if atleast one black ball is to be included in the draw is …….. .

Options

A. 129
B. 84
C. 64
D. none of these

Answer

To solve the problem of drawing three balls from a box containing 4 white balls, 5 black balls, and 2 red balls, with the condition that at least one black ball must be included, we will break the problem down into cases based on the number of black balls drawn.

### Step-by-Step Solution:

**Step 1: Identify the total number of balls.**
- White balls = 4
- Black balls = 5
- Red balls = 2
- Total balls = 4 + 5 + 2 = 11

**Step 2: Define the cases based on the number of black balls drawn.**
We will consider three cases:
1. Case 1: 1 black ball and 2 other balls.
2. Case 2: 2 black balls and 1 other ball.
3. Case 3: 3 black balls.

**Step 3: Calculate the number of ways for each case.**

**Case 1: 1 black ball and 2 other balls.**
- Choose 1 black ball from 5: \( \binom{5}{1} \)
- Choose 2 other balls from the remaining 6 (4 white + 2 red): \( \binom{6}{2} \)

Calculating:
\[ \binom{5}{1} = 5 \]
\[ \binom{6}{2} = \frac{6 \times 5}{2 \times 1} = 15 \]
Total ways for Case 1:
\[ 5 \times 15 = 75 \]

**Case 2: 2 black balls and 1 other ball.**
- Choose 2 black balls from 5: \( \binom{5}{2} \)
- Choose 1 other ball from the remaining 6: \( \binom{6}{1} \)

Calculating:
\[ \binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10 \]
\[ \binom{6}{1} = 6 \]
Total ways for Case 2:
\[ 10 \times 6 = 60 \]

**Case 3: 3 black balls.**
- Choose 3 black balls from 5: \( \binom{5}{3} \)

Calculating:
\[ \binom{5}{3} = \frac{5 \times 4}{2 \times 1} = 10 \]

**Step 4: Sum the total ways from all cases.**
Total ways = Case 1 + Case 2 + Case 3
\[ = 75 + 60 + 10 = 145 \]

### Final Answer:
The total number of ways to draw three balls from the box, ensuring at least one black ball is included, is **145**.

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