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Amy, Jean, Keith, Tom, Susan, and Dave have all been invited to a birthday party. They arrive randomly and each person arrives at a different time. In how many ways can they arrive? In how many ways can Jean arrive first and Keith last? Find the probability that Jean will arrive first and Keith will arrive last.

Respuesta :

Answer: a) 720, b) 24, c) 0.0333....

Step-by-step explanation:

Since we have given that

Number of people = 6

In how many ways can they arrive?

So, total possible ways would be

[tex]6!=720[/tex]

In how many ways can Jean arrive first and Keith last?

Since Jean take first place and keith takes the last place.

Remaining people = 4

so, number of ways would be

[tex]1!\times 4!\times 1!\\\\=24[/tex]

Find the probability that Jean will arrive first and Keith will arrive last.

So, probability would be

[tex]\dfrac{24}{720}=0.033...[/tex]

Hence, a) 720, b) 24, c) 0.0333....

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