Question

Find the number of ways in which 5 red balls, 4 balls of different sizes can be arranged in a row so that: -
(a) no two balls of the same colour come together.
(b) the balls of the same colour come together.

Answer 1

Hint: Consider the part (a) of the question and arrange all the red balls first with a space for a black ball between each of the two red balls. Arrange these red balls and among themselves using the relation $5!$ and blacks balls among themselves using the relation $4!$ to get the answer. Now, consider the part (b) of the question and arrange all the red balls side to each other with no space between them. In the next step arrange all the black balls together either at the place before these red balls or after them. Find the possible arrangements by arranging black balls among themselves and red balls among themselves to get the answer.

Complete step by step solution:
Here we have been provided with 5 red balls and 4 black balls and we are asked to find the possible arrangements possible in the given two conditions. Let us check them one by one.
(a) Here we have to make the arrangements such that two balls of the same colour may not come together. So arranging the red balls first at five places we get,
R _ R _ R _ R _ R
Now, to get the condition fulfilled we have to arrange 4 black balls at each of the four places vacant between each of the two red balls. So we get the arrangement as: -
R B R B R B R B R
Since the size of all the black balls and red balls are different, we can arrange the red balls among themselves and similarly the red balls among themselves.
$\Rightarrow $ Number of ways to arrange 5 red balls at 5 places = $5!$
$\Rightarrow $ Number of ways to arrange 4 black balls at 4 places = $4!$
Therefore total number of arrangements possible = $5!\times 4!=2880$

(b) Here we have to make the arrangements such that all the balls of the same colour come together. So arranging the red balls first at five places with no space between them we get,
R R R R R
Now, to get the condition fulfilled we have to arrange all the 4 black balls together either before the black balls or after them. So we get the arrangement as: -
R R R R R B B B B or B B B B R R R R R
Here also we can arrange the red balls among themselves and similarly the red balls among themselves.
$\Rightarrow $ Number of ways to arrange 5 red balls at 5 places = $5!$
$\Rightarrow $ Number of ways to arrange 4 black balls at 4 places = $4!$
Therefore total number of arrangements possible = $2\times 5!\times 4!=5760$

Note: Note that in part (b) of the solution we have multiplied $5!\times 4!$ with 2 in the end. This is because there were two possible arrangements of the groups of red balls and black balls either of these groups can be at the before or after the other, so mathematically it is equal to the relation ${}^{2}{{C}_{1}}$, i.e. the combination formula ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$.