Summer is Coming! Join the Game of Timers Competition to Win Epic Prizes. Registration is Open. Game starts Mon July 1st.

 It is currently 17 Jul 2019, 14:20

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

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

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 56277
How many positive four-digit integers have their digits in ascending  [#permalink]

### Show Tags

20 Dec 2017, 02:34
15
00:00

Difficulty:

95% (hard)

Question Stats:

32% (02:07) correct 68% (02:01) wrong based on 130 sessions

### HideShow timer Statistics

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

_________________
Math Expert
Joined: 02 Aug 2009
Posts: 7764
How many positive four-digit integers have their digits in ascending  [#permalink]

### Show Tags

01 Jan 2018, 07:34
6
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

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
_________________
##### General Discussion
Current Student
Joined: 18 Aug 2016
Posts: 617
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]

### Show Tags

20 Dec 2017, 06:23
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
_________________
We must try to achieve the best within us

Thanks
Luckisnoexcuse
Intern
Joined: 11 Apr 2014
Posts: 20
Re: How many positive four-digit integers have their digits in ascending  [#permalink]

### Show Tags

01 Jan 2018, 06:28
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.
Intern
Joined: 11 Apr 2014
Posts: 20
Re: How many positive four-digit integers have their digits in ascending  [#permalink]

### Show Tags

01 Jan 2018, 06:29
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.
Current Student
Joined: 18 Aug 2016
Posts: 617
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]

### Show Tags

01 Jan 2018, 07:07
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
Retired Moderator
Joined: 25 Feb 2013
Posts: 1197
Location: India
GPA: 3.82
Re: How many positive four-digit integers have their digits in ascending  [#permalink]

### Show Tags

01 Jan 2018, 07:44
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: 56277
Re: How many positive four-digit integers have their digits in ascending  [#permalink]

### Show Tags

24 Dec 2018, 05:01
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

_________________
Re: How many positive four-digit integers have their digits in ascending   [#permalink] 24 Dec 2018, 05:01
Display posts from previous: Sort by