Question

Given 5 different green toys, 4 different blue toys, and 3 different red toys, how many combinations of toys can be chosen taking at least one green and one blue toy?

• A. \(32 \times 16 \times 4\)
• B. \(31 \times 15 \times 4\)
• C. \(32 \times 16 \times 8\)
• D. \(31 \times 15 \times 8\)

Hint:

For "at least one" selections, use \(2^n - 1\). For "any number," use \(2^n\). Multiply across categories to get total combinations.

Answer 1

The Correct Option is D

We are to select toys such that: - At least one green (from 5), - At least one blue (from 4), - Any number of red (from 3). For green: Total non-empty subsets = \(2^5 - 1 = 31\) For blue: Total non-empty subsets = \(2^4 - 1 = 15\) For red: Any subset (including empty) = \(2^3 = 8\) So total combinations: \[ 31 \times 15 \times 8 \]