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

Question

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

Bunuel · Accepted Answer

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

Bunuel · Answer

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

Luckisnoexcuse · Answer

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

chetan2u · Answer

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

niks18 · Answer

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

Bunuel · Answer

Par of  GMAT CLUB'S New Year's Quantitative Challenge Set

PranjalJ · Answer

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

Posted from my mobile device

miag · Answer

Why can digits not repeat? Are 1122,2233,2344,1123 not considered valid numbers and if not, why? 
Want to make sure I understand this to keep in mind this constraints for future questions

Bunuel · Answer

Ascending order means the digits must increase from left to right, so for example, 1122 is not in ascending order because there’s no increase from the thousands to the hundreds digit and from the tens to the units digit.

Axwe7 · Answer

This wouldn't be correct since the units position cannot have 9 options. Because we're restricted to ascending order, the units position must be greater than or equal to 4 - so there's only 6 possibilites.

I.e., smallest arrangement is 1234.

I.e., smallest arrangement is 1234.

BinodBhai · Answer

This is a very good question indeed!

Vaishbab · Answer

He was talking about the 1000s digit

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

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