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Step 1: Understand the Problem
We have 10 white balls, 9 green balls, and 7 black balls. All balls of the same color are indistinguishable from one another. We need to find the number of ways to choose at least one ball from these sets of identical balls.
Step 2: Total Ways to Choose Zero or More Balls of Each Color
For each set of identical balls, the number of ways to choose any number of them (including the possibility of choosing none) is given by:
$$(\text{number of balls}) + 1$$
This is because for $n$ identical objects, the possible selections are 0, 1, 2, …, $n$, which totals $n+1$ ways.
Thus, for:
White balls: 10 → ways = $(10 + 1) = 11$
Green balls: 9 → ways = $(9 + 1) = 10$
Black balls: 7 → ways = $(7 + 1) = 8$
Step 3: Calculate the Number of Ways to Choose Balls (Including Zero Balls)
The total number of ways to choose white, green, and black balls (allowing zero of each) is the product:
$$(10 + 1) \times (9 + 1) \times (7 + 1) = 11 \times 10 \times 8$$
Compute this product:
$$11 \times 10 \times 8 = 880$$
Step 4: Subtract the Case of Choosing Zero Balls
Out of those $880$ ways, one way corresponds to selecting zero white AND zero green AND zero black, i.e., choosing no balls at all. Since the question asks for the number of ways to choose at least one ball, we subtract this 1 arrangement:
$$880 - 1 = 879$$
Step 5: Final Answer
Therefore, the number of ways to select at least one ball from the given sets of identical balls is:
$$879$$
