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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: 64216
How many positive four-digit integers have their digits in ascending  [#permalink]

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20 Dec 2017, 01:34
23
00:00

Difficulty:

95% (hard)

Question Stats:

32% (02:23) correct 68% (02:07) wrong based on 151 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

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Math Expert
Joined: 02 Aug 2009
Posts: 8601
How many positive four-digit integers have their digits in ascending  [#permalink]

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01 Jan 2018, 06:34
6
4
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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Joined: 18 Aug 2016
Posts: 588
Concentration: Strategy, Technology
GMAT 1: 630 Q47 V29
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Re: How many positive four-digit integers have their digits in ascending  [#permalink]

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20 Dec 2017, 05:23
2
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
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Luckisnoexcuse
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Joined: 11 Apr 2014
Posts: 19
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.
Current Student
Joined: 18 Aug 2016
Posts: 588
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"
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We must try to achieve the best within us

Thanks
Luckisnoexcuse
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Joined: 25 Feb 2013
Posts: 1124
Location: India
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
1
1
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 $$9_C_4=\frac{9!}{4!5!}=126$$
Math Expert
Joined: 02 Sep 2009
Posts: 64216
Re: How many positive four-digit integers have their digits in ascending  [#permalink]

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24 Dec 2018, 04:01
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

_________________
CEO
Joined: 03 Jun 2019
Posts: 2899
Location: India
GMAT 1: 690 Q50 V34
WE: Engineering (Transportation)
Re: How many positive four-digit integers have their digits in ascending  [#permalink]

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08 Oct 2019, 01:29
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

If we select 4 numbers out of 10 and deselect numbers with 0 as thousandth digit.
10C4 - 9C3 = 126

IMO D
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Kinshook Chaturvedi
Email: kinshook.chaturvedi@gmail.com
Intern
Joined: 10 Aug 2019
Posts: 20
Re: How many positive four-digit integers have their digits in ascending  [#permalink]

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11 Oct 2019, 05:38
Kinshook 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

If we select 4 numbers out of 10 and deselect numbers with 0 as thousandth digit.
10C4 - 9C3 = 126

IMO D

Why do we take combinations? I mean the order is important right(ascending) so shouldn't it be permutation?
Math Expert
Joined: 02 Sep 2009
Posts: 64216
Re: How many positive four-digit integers have their digits in ascending  [#permalink]

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02 Dec 2019, 23:25
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

Official Solution:

First of all, notice that 0 cannot be a digit of such numbers: 0 cannot be the first digit, since in this case a number will no longer be a four-digit number and become a three-digit number; and also it cannot be any other digit, since the digits must be in ascending order.

So, such numbers can have only 4 different digits out of 9 (all digits but 0). The number of groups of 4 different digits out of 9 is $$9C4=126$$ and since only one arrangement of any of such groups will be in ascending order then 126 is the final answer. For example, if the group of 4 we choose is {3, 5, 7, 1}, the only way to arrange this group of digits in ascending order is 1357.

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Re: How many positive four-digit integers have their digits in ascending   [#permalink] 02 Dec 2019, 23:25