AP EAMCET Engineering 22 Sep 2020 Shift 1 Paper

Number of ways to arrange 6 red balls are 6 !. Now, there are seven positions to place 6 black balls, such that no two black balls are togetherSo, number of ways to arrange 6 black balls are ‌7P6
Therefore the required number of arrangements =6!×‌7P6=6!×7!=7×6!×6!