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?
- A \(32 \times 16 \times 4\)
- B \(31 \times 15 \times 4\)
- C \(32 \times 16 \times 8\)
- D \(31 \times 15 \times 8\)
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…
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