**Concept: The question tests on the area of Counting**

**Solution:**

There are 8 identical blue bottles bottles and 7 identical green bottles. We want that 2 green bottles be together and so none of the remaining 5 green bottles can be together.

Let us take the 2 green bottles as 1 unit.

Let us first arrange the Blue bottles with the number of spaces between them indicated by numbers as shown below. In these spaces, we will first put the pair of green bottles and then place the remaining 5 bottles in the remaining spaces.

1 B1 2 B2 3 B3 4 B4 5 B5 6 B6 7 B7 8 B8 9

Since the 8 blue bottles are identical, there is only 1 way of arranging them.

Now for the pair of green bottles, there are 9 places for them to choose. The number of ways = 9C1 = 9 ways

For the last set of 5 bottles, they have 8 places to choose from = 8C5 = 8C3 = (8 * 7 * 6) / (3 * 2) = 56 ways

Therefore the total number of ways = 1 * 9 * 56 = 504

**(option c)**