How many permutations of simple events

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SUMMARY

The discussion focuses on calculating permutations of events involving 6 people. The first part confirms that the total arrangements of 6 individuals is 6! = 720. For the second part, when 3 specific individuals must stay together, they are treated as a single unit, resulting in 4! arrangements, which equals 24. The third part addresses the scenario where 2 specific individuals refuse to follow each other, requiring the use of complementary counting to find the total arrangements minus the arrangements where the two are together.

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Homework Statement



I am stuck on this problem relating to permutations:

The first part asks how many ways can 6 people be lined up. This answer, I believe is 6! = 720 ways.

The second part: If 3 specific persons, among 6, insist on following each other, how many ways are possible?

The third part: If 2 specific persons, among 6, refuse to follow each other, how many ways are possible?

The first part is easy, but I don't know how to do the second and third parts. What is the concept involved in solving the second and third parts?

Thanks for the help!
 
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For the second part, just think of the three-people clump as one person, so you have four people that you need to order.
 

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