Assuming the balls to be identical except for difference in colours, the number of ways in which one

Q: Wait, why is it $n+1$ and not just $n$?

Great question! If you have $n$ items, you can choose $1, 2, \dots, n$ items, which is $n$ ways. But you can also choose *zero* items. That 'zero' case is a valid selection in the counting process, so we add $1$ to account for it. For $10$ white balls, the choices are $0, 1, 2, \dots, 10$, which is $11$ options.

Q: Why do we multiply the choices instead of adding them?

Multiplication is for independent events occurring together. Choosing white balls doesn't restrict your choice of green balls. When you have independent tasks, you multiply the number of ways to perform each task to find the total number of combined outcomes.

Q: Why do we subtract $1$ at the end?

The product $(10+1)(9+1)(7+1)$ counts every possible combination, including the case where you pick $0$ white, $0$ green, and $0$ black balls. Since the problem demands 'one or more' balls, that 'all-zero' case is invalid, so we must subtract it.

Question

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 :

Options:

(A) 880

(B) 629

(C) 630

(D) 879

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in front of a table filled with colorful balls. You have $10$ white, $9$ green, and $7$ black balls. The challenge is to select one or more balls.

In combinatorics, the distinction between 'distinct' and 'identical' is everything. If the balls were distinct, we would be dealing with permutations and combinations of unique objects.

However, because the balls of the same color are indistinguishable, we are no longer asking *which* ball we pick, but *how many* of each color we choose.

### The Power of $n+1$

Let's focus on the white balls. We have $10$ identical white balls. If we want to select some white balls, we have a few options: we could pick zero, one, two, or all the way up to ten.

If you count these, you will see there are exactly $10+1 = 11$ possible outcomes. The $+1$ accounts for the choice of picking zero balls, which is a valid state in our counting process.

This logic applies to every color group:
*   For the $9$ green balls, we have $9+1 = 10$ choices.
*   For the $7$ black balls, we have $7+1 = 8$ choices.

### The Fundamental Principle of Counting

Since the selection of white balls is completely independent of the selection of green or black balls, we use the **Fundamental Principle of Counting**. This principle states that if one task can be done in $m$ ways and another in $n$ ways, the total number of ways to do both is $m 	imes n$.

Here, we have three independent tasks: selecting white, selecting green, and selecting black. The total number of combinations is the product of our choices:

$$11 	imes 10 	imes 8 = 880$$

### The Final Filter: The 'At Least One' Constraint

The product $880$ includes the case where we chose $0$ white, $0$ green, and $0$ black balls. In other words, it includes the scenario where we selected *no balls at all*.

The question explicitly asks for 'one or more' balls. Therefore, that 'all-zero' case is invalid. To get the correct answer, we subtract that one invalid case from our total:

$$880 - 1 = 879$$

### Conclusion

The final answer is **$879$**. The beauty of this problem lies in recognizing that when items are identical, the complexity of 'which' vanishes, leaving us with the elegant simplicity of 'how many'.

Remember the general formula for $p, q, r$ identical items: the number of ways to select at least one is:

$$(p+1)(q+1)(r+1) - 1$$

Keep this logic in your toolkit, and you will never be trapped by identical items again!

Animated Solution for Mathematics - Permutations and Combinations: 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 :

Select Answer:

Visualized Solution

Analyzing the Given Items

The Identical Items Principle

Choices for White Balls

Choices for Green Balls

Choices for Black Balls

Fundamental Principle of Counting

Calculating Total Combinations

The "At Least One" Constraint

Excluding the Empty Selection

Final Answer & General Formula

The Sigma Insight: Combinations and Selection

Analyzing the Setup

The Power of $n + 1$

The Fundamental Principle of Counting

The Final Filter

Conclusion

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