Question

There are $2$ identical white balls, $3$ identical red balls and $4$ green balls of different shades. The number of ways in which they can be arranged in a row so that at least one ball is separated from the balls of the same color is

• A. $6(7!-4!)$
• B. $7(6!-4!)$
• C. $8!-5!$
• D. $none$

Answer 1

The correct option is A $6(7!-4!)$

Here it can be solved by the total cases minus the number of cases in which no ball is separated.

$\because$ green balls are of different shades $\therefore$ total cases $=\frac{9!}{2!3!}$

Number of cases in all are together considering $2$ white ball as $1$ unit, $3$ red ball as other and $4$ green ball as $3rd$ is $3!$

However, intra permutation among $4$ green ball will take place,

$\therefore$ No. of ways $=\frac{9!}{2!3!}-3!\times 4!$