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In a box there are 90 slips, numbered from 1 to 90. John picks up four

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Manager
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Joined: 18 Jul 2019
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In a box there are 90 slips, numbered from 1 to 90. John picks up four  [#permalink]

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New post 25 Nov 2019, 07:08
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A
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Difficulty:

  65% (hard)

Question Stats:

33% (02:05) correct 67% (01:21) wrong based on 15 sessions

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In a box there are 90 slips, numbered from 1 to 90. John picks up four slips at random, one after the other, without replacement. What is the probability that the numbers on the slips, in the order he picks them up, are in ascending order?

(A)\(\frac{1}{6}\)

(B)\(\frac{1}{12}\)

(C)\(\frac{1}{24}\)

(D)\(\frac{1}{45}\)

(E) \(\frac{1}{49}\)
Spoiler: OA
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Re: In a box there are 90 slips, numbered from 1 to 90. John picks up four  [#permalink]

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New post 25 Nov 2019, 19:57
CaptainLevi wrote:
In a box there are 90 slips, numbered from 1 to 90. John picks up four slips at random, one after the other, without replacement. What is the probability that the numbers on the slips, in the order he picks them up, are in ascending order?

(A)\(\frac{1}{6}\)

(B)\(\frac{1}{12}\)

(C)\(\frac{1}{24}\)

(D)\(\frac{1}{45}\)

(E) \(\frac{1}{49}\)


All combination of selecting number is P(n,k) *permutation.There are exactly C(n,k) combination for the ascending one.So, it is C(n,k)/P(n,k) = 1/k!. For the current question it is 1/4! = 1/24
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Re: In a box there are 90 slips, numbered from 1 to 90. John picks up four  [#permalink]

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New post 26 Nov 2019, 12:52
Can someone explain this with more details, please?
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Re: In a box there are 90 slips, numbered from 1 to 90. John picks up four   [#permalink] 26 Nov 2019, 12:52
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