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 atleast one ball is separated from the balls of the same colour, is

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

Answer 1

Answer: (A)

$6(7!-4!)$

These are totally $9$ balls of which $2$ are identical of one kind, $3$ are a like of another kind $and$ $4$ district ones.

At least one ball of same color separated $=$ Total $-$ No ball of same color is separated

Total permutation $=\frac{9!}{2!3!}$

For no ball is separated : we consider all balls of same color as $1$ entity, so there are $3$ entities which can be placed in $3!$ ways.

The white and red balls are identical so they will be placed in $1$ way whereas green balls are different so they can be placed in $4!$ ways

$\Rightarrow Req=3!\times 4!$

At least one ball is separated $=\frac{9!}{2!3!}-3!4!$

$=\frac{9\times 8\times 7!}{2\times 6}-6\times 4!$

$=6(7!)-6(\times 4!)$

$=6(7!-4!).$

Hence, the answer is $6(7!-4!).$