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Re: Every night, Jon and his brothers randomly determine the sch [#permalink]
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Siddharth108 wrote:
Bunuel wrote:
Every night, Jon and his brothers randomly determine the schedule in which they will get to use the shower in the morning. How many distinct schedules are possible?

Sat there are n brothers including Jon. The number of distinct schedules would be n!.

(1) The probability that Jon will be first or last is 40%. The probability that Jon will be first is 1/n and the probability that he'll be last will also be 1/n. Thus we have that 1/n + 1/n = 0.4 --> n = 5. Sufficient.

(2) There are 48 ways Jon could be first or last. {Jon}{# of arrangements of other brothers} + {# of arrangements of other brothers}{Jon} = 48 --> (n-1)! + (n-1)! = 48 --> (n-1)! = 24 --> n = 5. Sufficient.

Answer: D.

Hope it's clear.



Hi Bunuel,

Could you please explain Statement (2)? How did (n-1)! + (n-1)! equal to 48, i.e. the number of ways Jon could be first OR last?

Thank you.


If Jon is first, then n-1 his brothers can be arranged in (n-1)! after him and if Jon is last, then n-1 his brothers can be arranged in (n-1)! before him.
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Re: Every night, Jon and his brothers randomly determine the sch [#permalink]
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Re: Every night, Jon and his brothers randomly determine the sch [#permalink]
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