M35-01

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:

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, note that 0 cannot be a digit in these numbers: 0 cannot be the first digit, as this would result in a three-digit number instead of a four-digit one; moreover, it cannot be any other digit, since the digits must be in ascending order and 0, being the smallest digit, cannot follow any other number.
 
Consequently, such numbers can only have 4 distinct digits out of the 9 possible options (excluding 0). The number of ways to choose 4 different digits from a set of 9 is  {9C4}=126 . Since there is only one ascending arrangement for each of these groups of 4 digits, 126 is the final answer. For instance, if we choose the group {3, 5, 7, 1}, the only way to arrange these digits in ascending order is 1357.

Answer: D

coreyander · Answer

I think this is a  high-quality  question and I agree with explanation.

Bitss · Answer

I think this is a  high-quality  question and I agree with explanation.

mallika29 · Answer

Why have you excluded the possibility tha digits cannot repeat? If we consider digits 1,2,2,3 the ascending order is 1,2,2,3. If we consider digits 1,1,1,1 then 1,1,1,1 is the ascending order. So repetition shoulde be allowed right?

Bunuel · Answer

Ascending order means that digit increase from left to right. So, 1223 is not in ascending order.

hydr01 · Answer

Why does 9*8*7*6 not give us the right answer?

bb · Answer

I think your calculation may give you something like the number of 4-digit integers that do not have repeating digits but it is not ascending (and I don't think you are accounting for a zero which can be in the spot 2, 3, 4 but not 1 in terms of unique, non-repeating numbers). Ascending means that the second digit one has to be greater than the first one and so on. The smallest one is 1234.

P.S. it is an excellent "wrong answer choice potentially" 3,024

Bunuel · Answer

This method calculates the number of 4-digit numbers that do not contain the digit 0 and have distinct digits. For instance, with this method you'll get all 4! = 24 possible numbers with digits (1, 2, 3, 4). However, since only one number from these 24 will have its digits in ascending order (namely, 1234), we need to adjust the count by dividing by the number of permutations of four digits, which is 4!. This yields the formula 9*8*7*6/4!, which gives the same answer of 126 as the official solution.

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

adityakaregamba · Answer

will the same be true for descending order as well ?

Bunuel · Answer

Yes. Out of 126 combinations only one will be in descending order.

anish0953 · Answer

in general if we arrange 1 ,2,2,,4,2,6 in ascending order , it comes as 1,2,2,2,4,6 . This is valid , and it was not where mentions for distinct , i might think the question is low quality or should at least me more specific ,, bb

Bunuel · Answer

The question is fine; you're interpreting it incorrectly. Ascending order means the digits must increase from left to right, so 1224 is not in ascending order because there’s no increase from the hundreds to the tens digit.

braddouglas10 · Answer

O great Bunuel, the man at the throne, do you have any more similar examples to this one?

Bunuel · Answer

BraydinWalters · Answer

When arranging a group of numbers, letters, people, etc. we should consider the combination and permutation formula.

Combination Formula =   -------------------  Number of Options!  -------------------------  
 ---------------------------  Number of Spots!  x  (Number of options - Number of Spots)!

Different from the Permutation Formula, we multiply the  (Number of Options - Number of Spots)!  by the  Number of Spots!   in the denominator. This will give us a larger denominator which will result in a smaller numerator and assure we are not double counting numbers.  
  For example, when we have 1234, we don't want to include 4321 in our total.

Number of Options!  = 9! (1, 2, 3, 4, 5, 6, 7, 8, 9)
 Number of Spots!  = 4! (the problem says thousands to unit digit so we have four spots)
 (Number of Options - Number of Spots)!  = (9-4)! = 5!

Back to our formula we have:

--- 9! ---   or  9 x 8 x 7 x 6  x 5 x 4 x 3 x 2 x 1  
4! x 5 ! ---- 4 x 3 x 2 x 1  x 5 x 4 x 3 x 2 x 1

= 126

D

braddouglas10 · Answer

Thank you, great Bunuel, the man at the throne

awaikar · Answer

I like the solution - it’s helpful.

Shivam2024 · Answer

I tried to solve this question using _ approach. The four digit number can be _ _ _ _. Thousands digit can't be 0 hence nine options for digits, similarly for hundreds, tens, and unit place number of options can be nine, eight, seven if repetition not allowed. Therefore maximum number of digits can be 9x9x8x7 = 4536. Now these four digits can be arranged in 4! ways and only one out of those 4! ways can all the digits be in ascending order. Thereby answer should be 4536/24 = 189. Where am I wrong conceptually.

M35-01

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