ExamBro
ExamBro
AP EAMCET · Maths · Permutation Combination

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?

  1. A \(32 \times 16 \times 4\)
  2. B \(31 \times 15 \times 4\)
  3. C \(32 \times 16 \times 8\)
  4. D \(31 \times 15 \times 8\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(31 \times 15 \times 8\)

Step-by-step Solution

Detailed explanation

Selecting atleast 1 green toy out of 5 can be done in \({ }^5 \mathrm{C}_1+{ }^5 \mathrm{C}_2+{ }^5 \mathrm{C}_3+{ }^5 \mathrm{C}_4+{ }^5 \mathrm{C}_5+=31\) ways. Selecting atleast 1 blue toy out of 4 can be done in…