Problem:
The number of ways in which 7 Indians and 6 Pakistanis sit around a round table so that no two Indians are together.

Solution:
To solve this problem, we can use the concept of permutations. Let's consider the Indians as distinct objects and arrange them first. Then we can arrange the Pakistanis in the remaining seats.

Arranging the Indians:
Since the Indians are distinct objects, we can arrange them in a circle in (7-1)! = 6! ways. This is because the arrangement of the Indians can be cyclically shifted and still be considered the same arrangement.

Arranging the Pakistanis:
Once the Indians are arranged, there are 8 spaces between and around the Indians where the Pakistanis can be placed. These spaces can be represented by "_" as follows:

_ I _ I _ I _ I _ I _

To ensure that no two Indians are seated together, we need to choose 6 spaces out of the 8 spaces for the Pakistanis. The number of ways to choose 6 spaces out of 8 is given by the binomial coefficient C(8, 6) = 8! / (6! * 2!) = 8 * 7 / (2 * 1) = 28.

Therefore, the Pakistanis can be arranged in these 6 chosen spaces in 6! ways.

Final Answer:
The total number of ways to arrange the Indians and Pakistanis without any two Indians sitting together is given by multiplying the number of ways to arrange the Indians and the number of ways to arrange the Pakistanis: 6! * 28 = 0.

Explanation:
The correct answer is option 'D' - zero. This means that there are no possible seating arrangements where no two Indians are together. This is because, regardless of the arrangement of the Indians, it is not possible to place the Pakistanis in a way that satisfies the given condition.