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# How many positive four-digit integers have their digits in ascending

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Math Expert
Joined: 02 Sep 2009
Posts: 43335

Kudos [?]: 139513 [0], given: 12794

How many positive four-digit integers have their digits in ascending [#permalink]

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20 Dec 2017, 01:34
Expert's post
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Difficulty:

95% (hard)

Question Stats:

31% (00:57) correct 69% (02:13) wrong based on 49 sessions

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GMAT CLUB'S CHALLENGE QUESTIONS:

How many positive four-digit integers have their digits in ascending order from thousands to units?

A. 6
B. 7
C. 84
D. 126
E. 210
[Reveal] Spoiler: OA

_________________

Kudos [?]: 139513 [0], given: 12794

Director
Joined: 18 Aug 2016
Posts: 627

Kudos [?]: 208 [2], given: 166

Concentration: Strategy, Technology
GMAT 1: 630 Q47 V29
GMAT 2: 740 Q51 V38
Re: How many positive four-digit integers have their digits in ascending [#permalink]

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20 Dec 2017, 05:23
2
KUDOS
Bunuel wrote:

GMAT CLUB'S CHALLENGE QUESTIONS:

How many positive four-digit integers have their digits in ascending order from thousands to units?

A. 6
B. 7
C. 84
D. 126
E. 210

First Digit - A
Second Digit - B
Third Digit - C
Fourth Digit - D

A can be 1-6
B can be 2-7
C can be 3-8
D can be 4-9
A=1; B = 2; C = 3 ; D = 6 options
A=1; B = 2; C = 4; D = 5 options
A = 1; B=2; C = 5; D = 4 options
A = 1; B=2; C = 6; D = 3 options
A = 1; B=2; C = 7; D = 2 options
A = 1; B=2; C = 8; D = 1 options
So When A =1 and B=2 then we have 21 options
WHen A = 1 and B=3 then we have 15 options
When A =1 and B=4 then we have 10 options
When A = 1 and B=5 then we have 6 options
When A = 1 and B=6 then we have 3 options
When A = 1 and B=7 then we have 1 option
So When A =1 we have 56 options
WHen A = 2 we have 35 options
When A = 3 we have 20 options
When A = 4 we have 10 options
When A = 5 we have 4 options
When A = 6 we have 1 option

Alltogether we have 56+35+20+10+4+1 = 126 options
D
Please correct me if i am wrong
_________________

We must try to achieve the best within us

Thanks
Luckisnoexcuse

Kudos [?]: 208 [2], given: 166

Intern
Joined: 11 Apr 2014
Posts: 10

Kudos [?]: [0], given: 31

Re: How many positive four-digit integers have their digits in ascending [#permalink]

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01 Jan 2018, 05:28
Re:
First Digit - A
Second Digit - B
Third Digit - C
Fourth Digit - D

A can be 1-6
B can be 2-7
C can be 3-8
D can be 4-9
A=1; B = 2; C = 3 ; D = 6 options
A=1; B = 2; C = 4; D = 5 options
A = 1; B=2; C = 5; D = 4 options
A = 1; B=2; C = 6; D = 3 options
A = 1; B=2; C = 7; D = 2 options
A = 1; B=2; C = 8; D = 1 options
So When A =1 and B=2 then we have 21 options
WHen A = 1 and B=3 then we have 15 options
When A =1 and B=4 then we have 10 options
When A = 1 and B=5 then we have 6 options
When A = 1 and B=6 then we have 3 options
When A = 1 and B=7 then we have 1 option
So When A =1 we have 56 options
WHen A = 2 we have 35 options
When A = 3 we have 20 options
When A = 4 we have 10 options
When A = 5 we have 4 options
When A = 6 we have 1 option

Alltogether we have 56+35+20+10+4+1 = 126 options
D
Please correct me if i am wrong
-----------------------------------------------------

in this solution the A is not being considered for '0'...

a clear and simple step would be 10c4.

Please clarify why 0 is left from the numbers.

Kudos [?]: [0], given: 31

Intern
Joined: 11 Apr 2014
Posts: 10

Kudos [?]: [0], given: 31

Re: How many positive four-digit integers have their digits in ascending [#permalink]

### Show Tags

01 Jan 2018, 05:29
Re:
First Digit - A
Second Digit - B
Third Digit - C
Fourth Digit - D

A can be 1-6
B can be 2-7
C can be 3-8
D can be 4-9
A=1; B = 2; C = 3 ; D = 6 options
A=1; B = 2; C = 4; D = 5 options
A = 1; B=2; C = 5; D = 4 options
A = 1; B=2; C = 6; D = 3 options
A = 1; B=2; C = 7; D = 2 options
A = 1; B=2; C = 8; D = 1 options
So When A =1 and B=2 then we have 21 options
WHen A = 1 and B=3 then we have 15 options
When A =1 and B=4 then we have 10 options
When A = 1 and B=5 then we have 6 options
When A = 1 and B=6 then we have 3 options
When A = 1 and B=7 then we have 1 option
So When A =1 we have 56 options
WHen A = 2 we have 35 options
When A = 3 we have 20 options
When A = 4 we have 10 options
When A = 5 we have 4 options
When A = 6 we have 1 option

Alltogether we have 56+35+20+10+4+1 = 126 options
D
Please correct me if i am wrong
-----------------------------------------------------

in this solution the A is not being considered for '0'...

a clear and simple step would be 10c4.

Please clarify why 0 is left from the numbers.

Kudos [?]: [0], given: 31

Director
Joined: 18 Aug 2016
Posts: 627

Kudos [?]: 208 [0], given: 166

Concentration: Strategy, Technology
GMAT 1: 630 Q47 V29
GMAT 2: 740 Q51 V38
Re: How many positive four-digit integers have their digits in ascending [#permalink]

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01 Jan 2018, 06:07
coolnaren wrote:
Re:
First Digit - A
Second Digit - B
Third Digit - C
Fourth Digit - D

A can be 1-6
B can be 2-7
C can be 3-8
D can be 4-9
A=1; B = 2; C = 3 ; D = 6 options
A=1; B = 2; C = 4; D = 5 options
A = 1; B=2; C = 5; D = 4 options
A = 1; B=2; C = 6; D = 3 options
A = 1; B=2; C = 7; D = 2 options
A = 1; B=2; C = 8; D = 1 options
So When A =1 and B=2 then we have 21 options
WHen A = 1 and B=3 then we have 15 options
When A =1 and B=4 then we have 10 options
When A = 1 and B=5 then we have 6 options
When A = 1 and B=6 then we have 3 options
When A = 1 and B=7 then we have 1 option
So When A =1 we have 56 options
WHen A = 2 we have 35 options
When A = 3 we have 20 options
When A = 4 we have 10 options
When A = 5 we have 4 options
When A = 6 we have 1 option

Alltogether we have 56+35+20+10+4+1 = 126 options
D
Please correct me if i am wrong
-----------------------------------------------------

in this solution the A is not being considered for '0'...

a clear and simple step would be 10c4.

Please clarify why 0 is left from the numbers.

Hey coolnaren ...
Can you give me an example of 4-digit no. in which digits are arranged in ascending order and has a "0"
_________________

We must try to achieve the best within us

Thanks
Luckisnoexcuse

Kudos [?]: 208 [0], given: 166

Math Expert
Joined: 02 Aug 2009
Posts: 5531

Kudos [?]: 6433 [2], given: 122

How many positive four-digit integers have their digits in ascending [#permalink]

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01 Jan 2018, 06:34
2
KUDOS
Expert's post
Bunuel wrote:

GMAT CLUB'S CHALLENGE QUESTIONS:

How many positive four-digit integers have their digits in ascending order from thousands to units?

A. 6
B. 7
C. 84
D. 126
E. 210

Hi..

an easier, short and logical way..

choose 4 out of 10... there can be ONLY 1 way the 4 can be arranged in ascending order, so selection is what we are looking for and NOT arrangements..
$$10C4 = \frac{10!}{6!4!} = \frac{10*9*8*7}{4*3*2} = 210$$

BUT it contains numbers with 0 in thousands digit making these 3-digit numbers
so $$9C3= \frac{9!}{6!3!} = 84$$

$$ans = 210-84=126$$

D
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Kudos [?]: 6433 [2], given: 122

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Joined: 25 Feb 2013
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GPA: 3.82
Re: How many positive four-digit integers have their digits in ascending [#permalink]

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01 Jan 2018, 06:44
chetan2u wrote:
Bunuel wrote:

GMAT CLUB'S CHALLENGE QUESTIONS:

How many positive four-digit integers have their digits in ascending order from thousands to units?

A. 6
B. 7
C. 84
D. 126
E. 210

Hi..

an easier, short and logical way..

choose 4 out of 10... there can be ONLY 1 way the 4 can be arranged in ascending order, so selection is what we are looking for and NOT arrangements..
$$10C4 = \frac{10!}{6!4!} = \frac{10*9*8*7}{4*3*2} = 210$$

BUT it contains numbers with 0 in thousands digit making these 3-digit numbers
so $$9C3= \frac{9!}{6!3!} = 84$$

$$ans = 210-84=126$$

D

Great explanation chetan2u

we can reduce 1 step here because $$0$$ cannot be part of any number with ascending digits so we simply need to select $$4$$ digits out of $$9$$ digits

Hence $$9_C_4=\frac{9!}{4!5!}=126$$

Kudos [?]: 381 [0], given: 42

Re: How many positive four-digit integers have their digits in ascending   [#permalink] 01 Jan 2018, 06:44
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