At a rehearsal dinner the night before a wedding, the bride and groom need to assign 10 people to two tables of five people. How many different seating arrangements are there?

Step 1 ：The problem can be broken down into two parts: Choosing 5 people out of 10 to sit at the first table and arranging the 5 people at each table.

Step 2 ：Choosing 5 people out of 10 to sit at the first table is a combination problem, because the order in which the people are chosen does not matter. The number of ways to do this is represented by the combination formula \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\), where \(n\) is the total number of people and \(k\) is the number of people to choose. Substituting \(n = 10\) and \(k = 5\), we get \(\binom{10}{5} = 252\).

Step 3 ：Arranging the 5 people at each table is a permutation problem, because the order in which the people are seated does matter. The number of ways to do this is represented by the permutation formula \(P(n) = n!\), where \(n\) is the number of people. Substituting \(n = 5\), we get \(P(5) = 5! = 120\).

Step 4 ：The total number of arrangements is the product of the number of ways to choose 5 people out of 10 and the number of ways to arrange 5 people at a table, squared (since there are two tables). So, the total number of arrangements is \(\binom{10}{5} \times P(5) \times P(5) = 252 \times 120 \times 120 = 3628800\).

Step 5 ：Final Answer: The total number of different seating arrangements is \(\boxed{3628800}\).