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27 Apr 2025, 04:41
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I don’t quite agree with the solution. if we use _ approach to solve this question then total number of four digit numbers that can be formed is 9x9x8x7. 9 for thousand place because 0 not allowed. 9 for hundred place because 0 allowed but thousand digit not allowed and so on. thus total 4536 numbers can be formed. Now divide that by 4! to get the numbers which are in ascending order. 4536/4! = 189.
27 Apr 2025, 09:07
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Shivam2024
Bunuel
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
I don’t quite agree with the solution. if we use _ approach to solve this question then total number of four digit numbers that can be formed is 9x9x8x7. 9 for thousand place because 0 not allowed. 9 for hundred place because 0 allowed but thousand digit not allowed and so on. thus total 4536 numbers can be formed. Now divide that by 4! to get the numbers which are in ascending order. 4536/4! = 189.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
Your method is wrong because when you do 9*9*8*7, you are counting all four-digit numbers with different digits, including numbers where 0 is one of the digits. But we already established that 0 cannot appear at all.
Also, when you divide by 4!, you are still including numbers where 0 is the first digit, which are not acceptable.
The original solution using 9C4 = 126 is correct. I suggest you review it and the full discussion more carefully.
27 Apr 2025, 10:32
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Okay I get it, maybe I missed the emphasis of the numbers being in ascending order. Thus 0 can not be part of the selection because it is the smallest integer from 0 - 9 as such no integer can precede it. Also if it acquires the thousands place the resultant number will ultimately be a three digit number. Can I use the logic (9x9x8x7)/4! had the question asked descending order?
Your method is wrong because when you do 9*9*8*7, you are counting all four-digit numbers with different digits, including numbers where 0 is one of the digits. But we already established that 0 cannot appear at all.
Also, when you divide by 4!, you are still including numbers where 0 is the first digit, which are not acceptable.
The original solution using 9C4 = 126 is correct. I suggest you review it and the full discussion more carefully.
Bunuel
Shivam2024
Bunuel
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
I don’t quite agree with the solution. if we use _ approach to solve this question then total number of four digit numbers that can be formed is 9x9x8x7. 9 for thousand place because 0 not allowed. 9 for hundred place because 0 allowed but thousand digit not allowed and so on. thus total 4536 numbers can be formed. Now divide that by 4! to get the numbers which are in ascending order. 4536/4! = 189.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
Your method is wrong because when you do 9*9*8*7, you are counting all four-digit numbers with different digits, including numbers where 0 is one of the digits. But we already established that 0 cannot appear at all.
Also, when you divide by 4!, you are still including numbers where 0 is the first digit, which are not acceptable.
The original solution using 9C4 = 126 is correct. I suggest you review it and the full discussion more carefully.
