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:

If at least one object must be selected from \(n\) distinct objects, the number of selections is \[ 2^n-1 \] because the empty set is excluded.

Answer 1

The Correct Option is D

Step 1: Understand the problem.
We have 5 green, 4 blue, 3 red toys. We need at least one green AND at least one blue toy. Red toys are optional (zero or more).

Step 2: Count selections from green toys (at least 1).
Each green toy is either selected or not: $2^5 = 32$ total subsets. Exclude the empty subset: $32-1=31$ ways to choose at least one green toy.

Step 3: Count selections from blue toys (at least 1).
Similarly, $2^4 = 16$ total subsets of blue toys. Exclude empty: $16-1=15$ ways.

Step 4: Count selections from red toys (any number, including none).
Red toys have no restriction. Each of the 3 red toys is in or out: $2^3=8$ ways.

Step 5: Apply the multiplication principle.
The three choices are independent of each other. Total combinations: \[ 31 \times 15 \times 8 \]

Step 6: State the answer.
\[ \boxed{31 \times 15 \times 8} \]