Maths-
General
Easy

Question

The sum of all the numbers that can be formed with the digits 2, 3, 4, 5 taken all at a time is (repetition is not allowed) :

  1. 93324    
  2. 66666    
  3. 84844    
  4. None of these    

hintHint:

 We start solving the problem by finding the total possibilities of getting numbers by fixing each digit in unit place. We then find the sum of all the numbers present in the unit place. Similarly, we multiply 10 for the sum of digits in tenth place, 100 for the sum of digits in tenth place and 1000 for the sum of digits in thousandth place. We then add all these sums to get the required answer.

The correct answer is: 93324


    Detailed Solution
    According to the problem, we need to find the sum of all the numbers that can be formed with the digits 2,3,4,5 taken all at a time.

    Let us first fix a number in a unit place and find the total number of words possible due on fixing this number.

    We need to arrange the remaining three places with three digits. We know that the number of ways of arranging n objects in n places is n! ways.
    So, we get  3! = 6  numbers on fixing the unit place with a particular digit.
    Now, let us find the sum of all digits.
    W e space g e t space s u m space a s space 2 space plus space 3 space plus space 4 space plus space 5 space equals space 14
    N o w comma space w e space g e t space a space s u m space o f space d i g i t s space i n space u n i t s space p l a c e space f o r space a l l space t h e space n u m b e r s space a s space 14 cross times 6 equals space 84.
    We use the same digits in ten, hundred and thousand places also. So, the sum of those digits will also be 84 but with the multiplication of its place value. 
    i.e., We multiply the sum of the digits in tenth place with 10, hundredth place with 100 and so on. We then add these sums.
    S o comma space w e space g e t space t h e space s u m space o f space a l l space t h e space n u m b e r s space t h a t space c a n space b e space f o r m e d space w i t h space t h e space d i g i t s space 2 comma 3 comma 4 comma 5 space t a k e n space a l l space a t space a space t i m e space i s space
left parenthesis 84 cross times 1000 right parenthesis plus left parenthesis 84 cross times 100 right parenthesis plus left parenthesis 84 cross times 10 right parenthesis plus left parenthesis 84 cross times 1 right parenthesis space equals space 93324
    Thus, the sum of all the numbers that can be formed with the digits 2, 3, 4, 5 taken all at a time is (repetition is not allowed) : 93324

     Alternatively, we can use the formula for the sum of numbers as
    left parenthesis n minus 1 right parenthesis factorial space cross times space left parenthesis S u m space o f space d i g i t s right parenthesis space cross times left parenthesis 11111...... space n t i m e s right parenthesis
    We can also solve this problem by writing all the possible numbers and finding the sum of them which will be time taking and make us confused. We should know that the value of the digits is determined by the place where they were present. We should check whether there is zero in the given digits and whether there are any repetitions present in the numbers. Similarly, we can expect problems to find the sum of numbers formed by these digits with repetition allowed.

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