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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.

  1. A \(6 ! \times 6 !\)
  2. B \(7 ! \times 6\) !
  3. C \(2 \times 6 ! \times 6 !\)
  4. D \(7 \times 6 ! \times 6 !\)
Verified Solution

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…