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Every night, Jon and his brothers randomly determine the sch

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Ronak275
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Every night, Jon and his brothers randomly determine the sch [#permalink] New post   Updated on: 03 Jul 2014, 12:07
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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?

(1) The probability that Jon will be first or last is 40%.

(2) There are 48 ways Jon could be first or last.
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Originally posted by Ronak275 on 03 Jul 2014, 12:02.
Last edited by Bunuel on 03 Jul 2014, 12:07, edited 1 time in total.
Renamed the topic and edited the question.
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Every night, Jon and his brothers randomly determine the sch [#permalink] New post   03 Jul 2014, 12:20
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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?

Say 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.
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Re: Every night, Jon and his brothers randomly determine the sch [#permalink] New post   06 Apr 2021, 04:47
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.
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Re: Every night, Jon and his brothers randomly determine the sch [#permalink] New post   06 Apr 2021, 11:24
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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] New post   05 Jun 2023, 06:05
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Re: Every night, Jon and his brothers randomly determine the sch [#permalink]
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