MultiplicationPrinciple Basic

Problem - 2476

Two Britons, three Americans, and six Chinese form a line:

  • How many different ways can the $11$ individuals line up?
  • If two people of the same nationality cannot stand next to each other, how many different ways can the $11$ individuals line up?

Without constraint, there are $\boxed{11!}$ ways to line up these people.

When people with same nationality cannot stand next to each other, we note that six Chinese must be separated by the five non-Chinese. Hence, the result is $\boxed{6!\times 5!}$ by considering the following steps:

  1. Line up the $6$ Chinese: there are $6!$ ways
  2. Line up the $5$ non-Chinese: there are $5!$ ways
  3. Merge these two lines by inserting one non-Chinese into the interval between two Chinese

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