AP EAMCET · Maths · Permutation Combination
Find the number of ways of arrangement 6 red balls and 6 black balls in a row such that no two black ball are together.
- A \(6 ! \times 6 !\)
- B \(7 ! \times 6\) !
- C \(2 \times 6 ! \times 6 !\)
- D \(7 \times 6 ! \times 6 !\)
Answer & Solution
Correct Answer
(D) \(7 \times 6 ! \times 6 !\)
Step-by-step Solution
Detailed explanation
Number of ways to arrange 6 red balls are 6 !. Now, there are seven positions to place \(6 \mathrm{black}\) balls, such that no two black balls are together So, number of ways to arrange 6 black balls are \({ }^7 P_6\). Therefore the required number of arrangements…
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