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Math Expert
Joined: 02 Sep 2009
Posts: 87568
How many positive four-digit integers have their digits in ascending
[#permalink]
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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
_________________
Math Expert
Joined: 02 Aug 2009
Posts: 10653
How many positive four-digit integers have their digits in ascending
[#permalink]
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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Math Expert
Joined: 02 Sep 2009
Posts: 87568
Re: How many positive four-digit integers have their digits in ascending [#permalink]
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:
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 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.
Answer: D
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Joined: 25 Feb 2013
Posts: 1004
Location: India
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Re: How many positive four-digit integers have their digits in ascending [#permalink]
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 \[9_C_4=\frac{9!}{4!5!}=126\]
Current Student
Joined: 18 Aug 2016
Posts: 549
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]
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: 18
Re: How many positive four-digit integers have their digits in
ascending [#permalink]
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: 549
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]
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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Thanks
Luckisnoexcuse
Math Expert
Joined: 02 Sep 2009
Posts: 87568
Re: How many positive four-digit integers have their digits in ascending [#permalink]
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
Par of GMAT CLUB'S New Year's Quantitative Challenge Set
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GMAT Club Legend
Joined: 03 Jun 2019
Posts: 4830
Location: India
GMAT 1: 690 Q50 V34
WE:Engineering [Transportation]
Re: How many positive four-digit integers have their digits in ascending [#permalink]
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
_________________
Kinshook Chaturvedi
Email:
Intern
Joined: 10 Aug 2019
Posts: 16
Re: How many positive four-digit integers have their digits in ascending
[#permalink]
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: 87568
Re: How many positive four-digit integers have their digits in ascending [#permalink]
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.
Answer: D
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Intern
Joined: 18 Apr 2020
Posts: 18
Location: Germany
GPA: 4
Re: How many positive four-digit integers have their digits in ascending
[#permalink]
I thought of doing this question using permutation so started with filling the 1000th position and we have 9[0 can not be used at 1000th position]options for that. While doing this I realized that answer options do not have big numbers so I can easily divide by 9 and check if I find a unique answer. I divided all options by 9 and only option D passed the test. It was a shortcut to do this question.
Posted from my mobile device
VP
Joined: 16 Jun 2021
Posts: 1072
Re: How many positive four-digit integers have their digits in ascending [#permalink]
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
Total ways of chosing 4 digits = 10C4 = 210
total ways of selecting with zero inthousand place =9C3 =84
=>210-83 = 126
Therefore IMO D
Manager
Joined: 01 Apr 2021
Posts: 60
Re: How many positive four-digit integers have their digits in ascending [#permalink]
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
I don't understand this method..isn't 8765[ which is in descending order] included in 10C4?
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Posts: 25253
Re: How many positive four-digit integers have their digits in ascending
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Re: How many positive four-digit integers have their digits in ascending [#permalink]
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