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C
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Difficulty:
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Question Stats:
32% (02:44) correct
68%
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wrong
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Find the sum of all possible 4 digit numbers that can be formed using all the digits of the number 1045.
A. 18000
B. 62220
C. 64440
D. 66660
E. None of these
Are You Up For the Challenge: 700 Level Questions
A. 18000
B. 62220
C. 64440
D. 66660
E. None of these
Are You Up For the Challenge: 700 Level Questions
25 Aug 2023, 00:32
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Bunuel
Method 1: Direct Method
How many such 4 digit numbers can be formed? 0 cannot be in thousands place (since it must be a 4 digit number)?
So we have 3 options for thousands place (1/4/5), then 3 remaining options for hundreds place, then 2 options for tens place and 1 option for units place. In all we get 3 * 3* 2 * 1 = 18 numbers
(using Basic Counting Principle discussed here: https://youtu.be/LFnLKx06EMU )
In these 18 numbers, 6 will have 1 in thousands place, 6 will have 4 in thousands place and 6 will have 5 in thousands place.
But how are the digits distributed in the other 3 places?
Consider units place. Say thousands digit is 1. So we are looking at numbers
1 _ _ _
We have 3 equivalent digits now - 0, 4 and 5 - so they can be arranged in 3! ways. Hence, in the 6 numbers, we will have 0 twice at units place, 4 twice at units place and 5 twice at units place.
1540
1450
1504
1054
1045
1405
The same will be true when the thousands digit is 4 or 5.
Hence in 18 numbers, 0 will appear 6 times in units place, and each of 1, 4 and 5 will appear 4 times.
The same will be true for tens and hundreds places too.
So sum of all units, tens and hundreds digits is 4*1 + 4*4 + 4*5 = 4 * (10) = 40
Sum of all thousands digits is 6*1 + 6*4 + 6*5 = 60
Sum of all numbers = 40 + 10*40 + 100*40 + 1000*60 = 64440
Method 2: Use the standard approach to find the sum of all such numbers and then subtract the sum of numbers with 0 in thousands place.
Assuming we get 4! numbers which is 24, each digit will appear at each place in 6 numbers.
So sum of units digit = 6(1 + 0 + 4 + 5) = 60
and same will be sum of tens, hundreds and thousands digits.
So sum of all numbers = 60 + 10*60 + 100*60 + 1000*60 = 60 * 1111 = 66660
From here, subtract the 3 digit numbers made using 1, 4 and 5 (such that 0 is in thousands place).
Sum of all these numbers = 20 * 111 = 2220
Sum of actual 4 digit numbers = 66660 - 2220 = 64440
Method 3: Formula method
If you know the formula that when we have to find the sum of all numbers that can be made by using all n digits without repetition, we can do it by using (n - 1)! * (Sum of digits) * 111... (n times)
Since we have 4 digits 1, 0, 4, 5, the sum of all numbers = (4-1)! * (1 + 0 + 4 + 5) * 1111 = 66660
Sum of all numbers with 3 digits 1, 4, 5 (and 0 in thousands place) = (3 - 1)! * (1 + 4 + 5) * 111 = 2220
Sum of only 4 digit numbers = 66660 - 2220 = 64440
Answer (C)
General Discussion
Archit3110
09 Dec 2021, 09:49
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Sum of all the numbers which can be formed by using n digits without repetition
(4-1)!*(1+0+4+5)*1111=
66660
Option D
Posted from my mobile device
(4-1)!*(1+0+4+5)*1111=
66660
Option D
Bunuel
Posted from my mobile device
