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# How many integers between 3000 and 4000 that have distinct

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Intern
Joined: 03 Aug 2013
Posts: 15
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Kudos [?]: 36 [2] , given: 46

How many integers between 3000 and 4000 that have distinct [#permalink]

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05 Sep 2013, 22:58
2
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Difficulty:

65% (hard)

Question Stats:

54% (02:49) correct 46% (02:09) wrong based on 113 sessions

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How many integers between 3000 and 4000 that have distinct digits and increase from left to right ?

(A) 20
(B) 48
(C) 60
(D) 120
(E) 600

My answer doesn't match with the OA so BUNUEL, NARENN, VERTITASKARISHMA, ZARRALOU, I am calling for your help
[Reveal] Spoiler: OA
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Joined: 31 Jan 2013
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Schools: ISB '15
WE: Consulting (Energy and Utilities)
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Kudos [?]: 49 [2] , given: 18

Re: How many integers between 3000 & 4000 having distinct digits [#permalink]

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05 Sep 2013, 23:54
2
KUDOS
Since the numbers must be distinct and increasing from left to right,

The only arrangements we could come-up with are:

345_ --> 4
346_ --> 3
347_ --> 2
348_ --> 1
356_ --> 3
357_ --> 2
358_ --> 1
367_ --> 2
368_ --> 1
378_ --> 1

Number of integers =
[Reveal] Spoiler:
20

/SW
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Kudos [?]: 12 [7] , given: 2

Re: How many integers between 3000 and 4000 that have distinct [#permalink]

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06 Sep 2013, 15:52
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You have 0 1 2 3 4 5 6 7 8 9; So eliminate 0 1 2 3; Remain S = {4, 5, 6, 7, 8, 9} (6 digits)
Because you need 3 distinct digits which increase from left to right, take 3 digits from S.
With 3 digits which increase from left to right : only one way to arrange; Ex: 468 (not 486)
It means: 6C3 = 6!/[3!x(6 - 3)!] = 6!/[3!x3!] = 4 x 5 x 6 / 3! = 4 x 5 = 20.
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Re: How many integers between 3000 and 4000 that have distinct [#permalink]

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24 Sep 2015, 00:35
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Re: How many integers between 3000 and 4000 that have distinct [#permalink]

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09 Sep 2016, 03:49
dngboy89nh wrote:
You have 0 1 2 3 4 5 6 7 8 9; So eliminate 0 1 2 3; Remain S = {4, 5, 6, 7, 8, 9} (6 digits)
Because you need 3 distinct digits which increase from left to right, take 3 digits from S.
With 3 digits which increase from left to right : only one way to arrange; Ex: 468 (not 486)
It means: 6C3 = 6!/[3!x(6 - 3)!] = 6!/[3!x3!] = 4 x 5 x 6 / 3! = 4 x 5 = 20.

For those who could not understand why we using Combination Formula here:

We have to pick 3 different digits from the Set S = {4, 5, 6, 7, 8, 9}

We are using Combination because we are only selecting the digits as there is only 1 way to arrange them i.e. increasing order.
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Re: How many integers between 3000 and 4000 that have distinct   [#permalink] 09 Sep 2016, 03:49
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