Question

In how many ways can 14 distinct red balls and 10 distinct blue balls be placed in a row such that 1) all red balls are adjacent, 2) all blue balls are adjacent, 3) no two red balls are adjacent?

Answer 1

a) As all the balls are distinct here, ( even though the colour could be same ) , therefore the total number ways in which these 24 balls could be placed here is given as the permutation of 24 items

= 24!

Therefore 24! is the number of ways that the 24 balls could be lined up adjascent to each other.

b) Number of ways to arrange 24 balls such that all 10 blue balls are adjascent to each other is computed as:

= Total permutation of 15 items that is 14 red balls and one whole grouo of 10 blue balls * Permutation of 10 blue balls within the group

= 15! * 10!

Therefore the total number of ways here are 15! * 10!

c) As the number of red balls here is 14 and blue balls here is 10, therefore it is not possible here to arrange these 24 balls such that no two red balls are adjascent. Therefore 0 ways are there here.