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Annie has 100 cards numbered 1 through 100. If she deals

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Intern
Joined: 01 Sep 2010
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Annie has 100 cards numbered 1 through 100. If she deals [#permalink]

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04 Oct 2010, 06:48
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Difficulty:

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Question Stats:

55% (01:41) correct 45% (01:52) wrong based on 117 sessions

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Annie has 100 cards numbered 1 through 100. If she deals five cards to Alex, without replacing any of them, what is the probability that Alex will get five consecutive numbers?

A. 95!/100!

B. 96!/100!

C. (95! X 5!)/100!

D. (96! X 5!)/100!

E. (97! X 4!)/100!
[Reveal] Spoiler: OA

Last edited by Bunuel on 09 Jul 2013, 08:46, edited 1 time in total.
RENAMED THE TOPIC.

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Re: Probability for consecutive numbers [#permalink]

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04 Oct 2010, 06:59
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Annie has 100 cards numbered 1 through 100. If she deals five cards to Alex, without replacing any of them, what is the probability that Alex will get five consecutive numbers?

A. 95!/100!

B. 96!/100!

C. (95! X 5!)/100!

D. (96! X 5!)/100!

E. (97! X 4!)/100!

I guess it's not necessary Alex to get 5 consecutive cards in ascending order, to have 5 consecutive at the end is good enough.

There are 96 consecutive numbers in 100: {1, 2, 3, 4, 5}, {2, 3, 4, 5, 6}, ..., {96, 97, 98, 99, 100};
Total ways to pick 5 cards out of 100 is $$C^5_{100}$$;

So, $$P=\frac{favorable \ outcomes}{total \ # \ of \ outcomes}=\frac{96}{C^5_{100}}=\frac{5!*96!}{100!}$$.

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Kudos [?]: 133160 [2], given: 12415

Intern
Joined: 04 Oct 2010
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Re: Probability for consecutive numbers [#permalink]

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04 Oct 2010, 07:05
Nice question! THanks for the explanation.

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Re: Probability for consecutive numbers [#permalink]

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04 Oct 2010, 08:23
excellent explanation... bunuel, you make tough questions seems so simple.... Awesome...
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Re: Probability for consecutive numbers [#permalink]

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06 Oct 2010, 02:18
Thanks Bunuel great explaination

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Re: Annie has 100 cards numbered 1 through 100. If she deals [#permalink]

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16 Oct 2017, 21:12
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Re: Annie has 100 cards numbered 1 through 100. If she deals   [#permalink] 16 Oct 2017, 21:12
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