Math problem:

Suppose we randomly seat 7 Canadians and 5 Americans in a row. Find the probability that all the Canadians are seated togetheror all the Americans are seated together?

Below is my solution for this question, please verify whether it is correct or not. If it is incorrect please provide guidance on how to correctly solve it.

My Solution

All possible arrangements of 7 Canadians and 5 Americans can be calculated as;

12!/7!5! =792 possible arrangements.

To find the probability of 7 Canadians seated together, it can be treated as 1 object. In total

we are arranging 6 objects.

6!/5!1! = 6 distinct arrangements

P(Canadians are seated together) = 6/792

To find the probability of 5 Americans seated together, it can be treated as 1 object. In total

we are arranging 8 objects.

8!/7!1! = 8 distinct arrangements

P(Americans are seated together) = 8 /792

P(Canadians OR Americans are seated together) = (6/792) + (8/792) = 14/792

= 0.0176

The probability that all the Canadians are seated together or all the Americans are seated together is 0.0176