Question

we have 9 balls of colours red, green and blue, 3 balls of each color (balls of the same color are identical). we want to place them in the sequence in such a way that no two blue balls are next to each other. how many ways do we have to do it?

Answer 1

To arrange the balls in the given sequence without having two blue balls next to each other, there are 840 different ways to do it.

Step-by-step explanation:

- If we start with a blue ball, we have 8 remaining spaces to place the other blue balls. The first blue ball can be placed in 8 different ways. Once the first blue ball is placed, there are 7 spaces left. The second blue ball can be placed in 7 different ways. Finally, the third blue ball can be placed in 6 different ways.

- If we start with a red or green ball, we have 9 remaining spaces to place the blue balls. The first ball can be placed in 9 different ways. Then, we have 8 remaining spaces to place the second blue ball, which can be done in 8 different ways. Finally, we have 7 remaining spaces to place the third blue ball, which can be done in 7 different ways.

Since we want to avoid having two blue balls next to each other, we need to consider the cases where the first two blue balls are consecutive. For the first case, we have 8 x 7 x 6 = 336 possible arrangements. For the second case, we have 9 x 8 x 7 = 504 possible arrangements. Therefore, the total number of ways to arrange the balls is 336 + 504 = 840.