Final answer:
To count the number of ways to seat 20 people in 10 couples with each couple adjacent, first consider each couple as a unit and arrange the 10 units, then multiply the arrangements within each couple, giving 10! × 2^10 total arrangements.
Step-by-step explanation:
The question deals with the concept of counting and arranging items in permutations, which is a common problem in combinatorics, a branch of mathematics. Since each couple must sit together and there are 10 couples, we can temporarily treat each couple as a single unit.
This means that we effectively have 10 units to arrange in a row, which can be done in 10! ways, as we can arrange 10 unique items in 10 factorial ways. Once we have fixed the positions of these 10 'couple' units, we must consider the arrangements within each couple. Since there are 2 people in each couple and they can be arranged in 2! ways, for 10 couples that gives us an additional 210 arrangements.
Therefore, the total number of ways we can seat these 20 people in 20 seats, ensuring each couple sits together, is the product of these two figures: 10! (arrangements of the couples as units) multiplied by 210 (arrangements within each couple).
The answer is 10! × 210 possible seating arrangements that satisfy the condition of the problem.