Problem:
Find the number of ways in which 2 Indians, 3 Americans, 3 Italians, and 4 Frenchmen can be seated on a circle if people of the same nationality sit together.

Solution:

Step 1:
First, let's consider each nationality as a single entity. We will treat the Indians as one group, the Americans as another group, the Italians as another group, and the Frenchmen as the final group. This simplifies the problem as we only have four entities to arrange.

Step 2:
Now, let's fix the position of the Indians. Since there are 2 Indians, they can be arranged in 2! (2 factorial) ways.

Step 3:
Next, let's fix the position of the Americans. Since there are 3 Americans, they can be arranged in 3! ways.

Step 4:
Similarly, we can fix the position of the Italians in 3! ways and the Frenchmen in 4! ways.

Step 5:
Now that we have fixed the positions of each nationality, we can arrange these four groups on the circle. Since the circle is symmetrical, we divide the total number of arrangements by 4 to avoid overcounting. This is because rotating the circle would result in the same arrangement.

Step 6:
To calculate the final answer, we multiply the number of ways to arrange each group together (2! * 3! * 3! * 4!) and divide it by 4.

Step 7:
Using the formula, we can calculate the number of ways as follows:
(2! * 3! * 3! * 4!) / 4 = 2 * 6 * 6 * 24 / 4 = 288

Answer:
Therefore, there are 288 different ways in which 2 Indians, 3 Americans, 3 Italians, and 4 Frenchmen can be seated on a circle if people of the same nationality sit together.