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Wed Feb 10, 2021 4:00 am
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
39% (01:56) correct
61%
(02:03)
wrong
based on 106
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Time
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Number of positive integers which have no two digits having the same value with sum of their digits being 45, is
(A) 10!
(B) 9*9!
(C) 17*8!
(D) 9!
(E) 8!
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
(A) 10!
(B) 9*9!
(C) 17*8!
(D) 9!
(E) 8!
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
Wed Feb 10, 2021 4:36 am
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If no two digits should have the same value, we can use each digit from 0 to 9 once only. Only sum of all these digits equals to 45. So our numbers should contain a) all digits from 0 to 9 or b) all digits from 1 to 9.
a) 10 digit number: We can form 10!different numbers using the digits from 0-10.
However, first digit cannot be 0 and 10! includes numbers whose digit is 0.
In order to exlude these numbers we subtract the number of numbers whose 1st digit is 0.
So actually we can form 10!-9! different numbers from these 10 digits.
b) 9-digit numbers: We can make 9! different numbers using these 9 digits.
In total we can form 10!-9!+9!=10! different numbers.
IMO answer is A)
a) 10 digit number: We can form 10!different numbers using the digits from 0-10.
However, first digit cannot be 0 and 10! includes numbers whose digit is 0.
In order to exlude these numbers we subtract the number of numbers whose 1st digit is 0.
So actually we can form 10!-9! different numbers from these 10 digits.
b) 9-digit numbers: We can make 9! different numbers using these 9 digits.
In total we can form 10!-9!+9!=10! different numbers.
IMO answer is A)
Thu Jul 30, 2026 10:28 pm
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This is the first time I am giving a solution to a problem, pls correct me if I am wrong
as the digits must not repeat and also the sum of digits is 45,
If we add the digits 0-9, the sum is 45, thus all the digits can be used therefore, no of ways of forming digits-
For first digit- 10 ways
For 2nd-9 ways
For 3rd-8 ways
For 4th- 7 thus upto last will be 1, multiplying all give us 10¡ as answer.
as the digits must not repeat and also the sum of digits is 45,
If we add the digits 0-9, the sum is 45, thus all the digits can be used therefore, no of ways of forming digits-
For first digit- 10 ways
For 2nd-9 ways
For 3rd-8 ways
For 4th- 7 thus upto last will be 1, multiplying all give us 10¡ as answer.
Fri Jul 31, 2026 1:39 am
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Hi OnePieceIsReal,
Welcome, and since you asked to be corrected, there is a small mistake here: your answer is right, but the road you took to it isn't. That deserves more of your attention than a wrong answer would, because a method that accidentally works is a method you'll trust again.
Your first move is the real insight, though. Since 0+1+2+...+9 = 45, the digits have to sum to the whole set. Hold onto that.
The correct route, in two steps
Step 1 - find every digit set that sums to 45. There are two, not one. The full set {0,1,...,9} sums to 45, and so does {1,2,...,9}, because dropping the 0 costs nothing. Remove any other digit and the sum falls short. So two families of numbers qualify.
Step 2 - count each family, and never let 0 lead.
- Using all ten digits: total arrangements are 10!, but the ones with 0 in front aren't 10-digit numbers at all. Fix 0 in the lead and arrange the other nine: 9! of those. Valid count = 10! - 9!.
- Using {1,...,9}: no 0 in the pool, so every arrangement is a legal 9-digit number: 9!.
Total = (10! - 9!) + 9! = 10!, choice A.
Why your two mistakes hid each other
You counted the leading-zero strings you should have discarded, and you skipped the 9-digit family you should have added. Each group has exactly 9! members, and that's no coincidence: take any discarded string, delete its leading 0, and what's left is one of those 9-digit numbers. One-to-one, both directions. Your two errors were the same size, so they cancelled and the total still read 10!.
The habit to build
Whenever 0 sits in your digit pool, one question comes first, every time: can 0 take the lead? Quick test - how many 2-digit numbers use each of {0, 1} exactly once? It looks like 2x1 = 2, but 01 is just 1. Only 10 qualifies, so the answer is 1.
Your instinct on the digit sum was excellent. Now make that leading-zero check automatic.
Answer: A
Welcome, and since you asked to be corrected, there is a small mistake here: your answer is right, but the road you took to it isn't. That deserves more of your attention than a wrong answer would, because a method that accidentally works is a method you'll trust again.
Your first move is the real insight, though. Since 0+1+2+...+9 = 45, the digits have to sum to the whole set. Hold onto that.
The correct route, in two steps
Step 1 - find every digit set that sums to 45. There are two, not one. The full set {0,1,...,9} sums to 45, and so does {1,2,...,9}, because dropping the 0 costs nothing. Remove any other digit and the sum falls short. So two families of numbers qualify.
Step 2 - count each family, and never let 0 lead.
- Using all ten digits: total arrangements are 10!, but the ones with 0 in front aren't 10-digit numbers at all. Fix 0 in the lead and arrange the other nine: 9! of those. Valid count = 10! - 9!.
- Using {1,...,9}: no 0 in the pool, so every arrangement is a legal 9-digit number: 9!.
Total = (10! - 9!) + 9! = 10!, choice A.
Why your two mistakes hid each other
You counted the leading-zero strings you should have discarded, and you skipped the 9-digit family you should have added. Each group has exactly 9! members, and that's no coincidence: take any discarded string, delete its leading 0, and what's left is one of those 9-digit numbers. One-to-one, both directions. Your two errors were the same size, so they cancelled and the total still read 10!.
The habit to build
Whenever 0 sits in your digit pool, one question comes first, every time: can 0 take the lead? Quick test - how many 2-digit numbers use each of {0, 1} exactly once? It looks like 2x1 = 2, but 01 is just 1. Only 10 qualifies, so the answer is 1.
Your instinct on the digit sum was excellent. Now make that leading-zero check automatic.
Answer: A
OnePieceIsReal
