Find the number of ways of arranging 6 red ba | vedclass

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(D) First,arrange the $6$ red balls in a row. The number of ways to arrange $6$ identical red balls is $1$ (since they are identical),but if we treat

Vedclass · Accepted Answer

$7 	imes 6! 	imes 6!$

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(D) First,arrange the $6$ red balls in a row. The number of ways to arrange $6$ identical red balls is $1$ (since they are identical),but if we treat them as distinct for the sake of permutation logic,it is $6!$. However,usually,balls of the same color are considered identical. Assuming the balls are distinct,the arrangement is $6!$. There are $7$ possible gaps created by $6$ red balls (including the ends) to place the $6$ black balls: $\_ R \_ R \_ R \_ R \_ R \_ R \_$. The number of ways to choose $6$ positions out of $7$ is $\binom{7}{6} = 7$. The number of ways to arrange $6$ black balls in these chosen positions is $6!$. Thus,the total number of arrangements is $6! 	imes \binom{7}{6} 	imes 6! = 6! 	imes 7 	imes 6! = 7 	imes 6! 	imes 6!$.

Find the number of ways of arranging $6$ red balls and $6$ black balls in a row such that no two black balls are together.

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