There are 12 balls of which 4 are red, 3 black and 5 white: In how many ways can you arrange the balls so that no two white balls may occupy consecutive positions if All balls are considered to be different.

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Solution

All balls are considered to be different . As above arrange 7 non - White balls in 7! ways and in the 8 gaps arrange 5 different white balls in
$^{8} P_{3} = \frac{8!}{3!}$ ways. Hence by the fundamental theorem
$\frac{7! 8!}{3!}$ is required numbers ways.

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There are 12 balls of which 4 are red, 3 black and 5 white: In how many ways can you arrange the balls so that no two white balls may occupy consecutive positions if All balls are considered to be different.

All balls are considered to be different . As above arrange 7 non - White balls in 7! ways and in the 8 gaps arrange 5 different white balls in 8P3=8!3! ways. Hence by the fundamental theorem 7!8!3! is required numbers ways.

All balls are considered to be different . As above arrange 7 non - White balls in 7! ways and in the 8 gaps arrange 5 different white balls in
$^{8} P_{3} = \frac{8!}{3!}$ ways. Hence by the fundamental theorem
$\frac{7! 8!}{3!}$ is required numbers ways.