A panel for a political forum is made up of 11 people from three parties, all seated in a row, The panel consists of 4 Greet Party members, 1 Labour Party member, and 6 Republicans. In how many distinct orders can they be seated if two people of the same party are considered identical (not distinct)?

Step 1 ：The problem is asking for the number of distinct arrangements of the 11 people, where people from the same party are considered identical. This is a permutation problem involving identical objects.

Step 2 ：The formula for the number of permutations of n objects, where there are n1 of one type, n2 of another type, and so on, is \(\frac{n!}{n1! * n2! * ...}\).

Step 3 ：In this case, n = 11 (the total number of people), n1 = 4 (the number of Green Party members), n2 = 1 (the number of Labour Party members), and n3 = 6 (the number of Republicans).

Step 4 ：So, we need to calculate \(\frac{11!}{4! * 1! * 6!}\).

Step 5 ：Final Answer: The number of distinct orders in which the panel members can be seated is \(\boxed{2310}\).