Question

Find the number of ways in which $5$ red balls, $4$ black balls of different sizes can be arranged in a row so that
$\left(i\right)$ no two balls of the same colour come together
$\left(ii\right)$ the balls of the same colour come together.

Answer 1

We can set up the position of the red balls in this way

$i)$$BRBRBRBRB$
Where $R$ denotes the places where we can keep the red balls and $B$ denotes the places where we can keep the black balls.

Now, the red balls can be arranged within themselves in $4!$ ways.

In the $5$ gaps that are available between the red balls, we have to arrange to black balls. They can be arranged in ${}^5{C}_5$ ways which is $5!.$

Hence,

Total possible ways of arranging balls such that

no two balls of the same color come together $4!\times 5!=2880.$

We can set up the position of the red balls in this way

$ii)$There are two possibilities of arranging balls in such a way that$BBBBBRRRR$ or $RRRRBBBBB$
Where $R$ denotes the places where we can keep the red balls and $B$ denotes the places where we can keep the black balls.

Now, the red balls can be arranged within themselves in $4!$ ways.

In the $5$ gaps that are available between the red balls, we have to arrange to black balls. They can be arranged in ${}^5{C}_5$ ways which is $5!.$

Hence,

Total possible ways of arranging balls such that

no two balls of the same color come together $2\times 4!\times 5!=5760.$

Hence, this is the answer.