There are 7 identical white balls and 3 ident | vedclass

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(C) To solve this,we use the gap method.Since the $7$ white balls are identical,they can be arranged in only $1$ way.To ensure no two black balls are

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$56$

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(C) To solve this,we use the gap method.Since the $7$ white balls are identical,they can be arranged in only $1$ way.To ensure no two black balls are adjacent,we place the $3$ black balls in the gaps created by the $7$ white balls.Representing white balls as $W$,the arrangement is: $\_ W \_ W \_ W \_ W \_ W \_ W \_ W \_$.There are $8$ available slots for the $3$ identical black balls.The number of ways to choose $3$ slots out of $8$ is given by ${}^8C_3$.${}^8C_3 = \frac{8 	imes 7 	imes 6}{3 	imes 2 	imes 1} = 56$.Thus,the total number of distinguishable arrangements is $56$.

There are $7$ identical white balls and $3$ identical black balls. The number of distinguishable arrangements in a row of all the balls,so that no two black balls are adjacent is

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