3. Find the number of ways in which red balls, 4 black balls of different sizes can be arranged in a
row so that (i) no two balls of the same colour come together
Answers
Step-by-step explanation:
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!×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 thatBBBBBRRRR 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 2×4!×5!=5760.
Hence, this is the answer.