How many ways can 5 identical red balls and 4 identical blue balls be arranged in a row such that no two blue balls are adjacent?

The problem can be solved using the concept of permutations and combinations.

Step 1: Arrange the red balls

First, arrange the 5 identical red balls in a row. Since they are identical, there is only 1 way to do this.

R R R R R

Step 2: Identify the spaces between and at the ends of the red balls

There are 6 spaces where the blue balls can be placed (indicated by the underscores):

_ R _ R _ R _ R _ R _

Step 3: Place the blue balls

We have 4 identical blue balls and 6 spaces. We need to choose 4 spaces out of 6 to place the blue balls. Since the blue balls are identical, this is a combination problem.

The number of ways to choose 4 spaces out of 6 is given by the combination formula C(n, r) = n! / [r!(n-r)!], where n is the total number of items, r is the number of items to choose, and "!" denotes factorial.

So, the number of ways to place the blue balls is C(6, 4) = 6! / [4!(6-4)!] = 15.

Therefore, there are 15 ways to arrange 5 identical red balls and 4 identical blue balls in a row such that no two blue balls are adjacent.