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

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

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New post 20 Dec 2017, 02:34
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How many positive four-digit integers have their digits in ascending  [#permalink]

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

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

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

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New post 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.
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Joined: 18 Aug 2016
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Re: How many positive four-digit integers have their digits in ascending  [#permalink]

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

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New post 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 :thumbup:

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 \(9_C_4=\frac{9!}{4!5!}=126\)
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Re: How many positive four-digit integers have their digits in ascending  [#permalink]

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New post 24 Dec 2018, 05:01
GMAT Club Bot
Re: How many positive four-digit integers have their digits in ascending   [#permalink] 24 Dec 2018, 05:01
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