A box contains 2 black, 4 white, and 3 red balls. One ball is drawn at random from the box and kept aside. From the remaining balls in the box, another ball is drawn at random and kept aside the first. This process is repeated till all the balls are drawn front the box. The probability that the balls drawn are in the sequence of 2 black, 4 white, and 3 red is `1//1260` b. `1//7560` c. `1//126` d. none of these

Options

A. `1//1260`
B. `1//7560`
C. `1//126`
D. None of these

Explanation

To solve the problem, we need to find the probability of drawing the balls in the specific sequence of 2 black, 4 white, and 3 red from a box containing 2 black, 4 white, and 3 red balls.

### Step-by-Step Solution:

1. **Identify Total Balls**:
- We have 2 black, 4 white, and 3 red balls.
- Total number of balls = 2 + 4 + 3 = 9.

2. **Probability of Drawing the First Black Ball**:
- The probability of drawing a black ball first is the number of black balls divided by the total number of balls.
- Probability = Number of black balls / Total balls = 2 / 9.

3. **Probability of Drawing the Second Black Ball**:
- After drawing the first black ball, there is now 1 black ball left and a total of 8 balls remaining.
- Probability = Number of remaining black balls / Total remaining balls = 1 / 8.

4. **Probability of Drawing the First White Ball**:
- Now, we have 4 white balls and 7 balls remaining.
- Probability = Number of white balls / Total remaining balls = 4 / 7.

5. **Probability of Drawing the Second White Ball**:
- After drawing the first white ball, we have 3 white balls left and 6 balls remaining.
- Probability = Number of remaining white balls / Total remaining balls = 3 / 6 = 1 / 2.

6. **Probability of Drawing the Third White Ball**:
- After drawing the second white ball, we have 2 white balls left and 5 balls remaining.
- Probability = Number of remaining white balls / Total remaining balls = 2 / 5.

7. **Probability of Drawing the Fourth White Ball**:
- After drawing the third white ball, we have 1 white ball left and 4 balls remaining.
- Probability = Number of remaining white balls / Total remaining balls = 1 / 4.

8. **Probability of Drawing the First Red Ball**:
- Now, we have 3 red balls and 3 balls remaining.
- Probability = Number of red balls / Total remaining balls = 3 / 3 = 1.

9. **Probability of Drawing the Second Red Ball**:
- After drawing the first red ball, we have 2 red balls left and 2 balls remaining.
- Probability = Number of remaining red balls / Total remaining balls = 2 / 2 = 1.

10. **Probability of Drawing the Third Red Ball**:
- After drawing the second red ball, we have 1 red ball left and 1 ball remaining.
- Probability = Number of remaining red balls / Total remaining balls = 1 / 1 = 1.

11. **Calculate the Total Probability**:
- The total probability of drawing the balls in the sequence of 2 black, 4 white, and 3 red is the product of all the individual probabilities:
\[ P = \left(\frac{2}{9}\right) \times \left(\frac{1}{8}\right) \times \left(\frac{4}{7}\right) \times \left(\frac{3}{6}\right) \times \left(\frac{2}{5}\right) \times \left(\frac{1}{4}\right) \times \left(\frac{3}{3}\right) \times \left(\frac{2}{2}\right) \times \left(\frac{1}{1}\right) \]

12. **Simplifying the Expression**:
- This simplifies to:
\[ P = \frac{2 \times 1 \times 4 \times 3 \times 2 \times 1 \times 3 \times 2 \times 1}{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} \]
- The numerator becomes \(2 \times 4 \times 3 \times 2 \times 3 \times 2 = 144\).
- The denominator is \(9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 362880\).

13. **Final Calculation**:
- Therefore, the probability is:
\[ P = \frac{144}{362880} = \frac{1}{2520} \]

14. **Check the Options**:
- The closest option is \( \frac{1}{1260} \), which is incorrect. The correct answer should be \( \frac{1}{2520} \).