Assuming the balls to be identical except for difference in colours, the number of ways in which one or more balls can be selected from 10 white, 9 green and 7 black balls is (1) 880 (2) 629 (3) 630 (4) 879

Options

A. 630
B. 879
C. 880
D. 629

Explanation

To solve the problem of selecting one or more balls from a collection of 10 white, 9 green, and 7 black balls, we can use the principle of combinations for identical objects of different types.

### Step-by-Step Solution:

1. **Identify the Variables**:
- Let \( P \) be the number of white balls = 10
- Let \( Q \) be the number of green balls = 9
- Let \( R \) be the number of black balls = 7

2. **Apply the Formula**:
The formula for the number of ways to select one or more objects from groups of identical objects is given by:
\[ (P + 1)(Q + 1)(R + 1) - 1 \]
Here, the "+1" accounts for the option of selecting none of the balls of that color, and we subtract 1 at the end to exclude the case where no balls are selected.

3. **Substitute the Values**:
Substitute the values of \( P \), \( Q \), and \( R \) into the formula:
\[ (10 + 1)(9 + 1)(7 + 1) - 1 \]
This simplifies to:
\[ (11)(10)(8) - 1 \]

4. **Calculate the Product**:
Now, calculate the product:
\[ 11 \times 10 = 110 \]
\[ 110 \times 8 = 880 \]

5. **Subtract 1**:
Finally, subtract 1 from the total:
\[ 880 - 1 = 879 \]

### Final Answer:
The number of ways in which one or more balls can be selected from the given collection is **879**.