Find Sum of four digit Number

Question

Find the sum of all four digit numbers that can be formed by the digits {0,1,2,3,4}?

a)183450

b)259980

c)266640

d)6600

Answer 1

Option (b)  is correct.

Method 1(Using Expectation):

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No. of 4 digit no possible using {0,1,2,3,4}   =  4*4*3*2  =  96

Now expected value of first digit in any given no. = (1 +2 +3 +4)/4 =  5/2

Now sum of all digits : 0 +1 + 2 +3 + 4 = 10

= sum of all five expected digits = E[1st digit] + 4* E[other digit except 1st] = 5/2 + 4*E[other digit except 1st] 

Hence ,4*E[other digit except 1st]  = 10 - 5/2  = 15/2 => E[other digit except 1st]  = 15/8

Now since there are 96 numbers, Hence total sum = 96*[5/2 *10³ + 15/8 * 10²  + 15/8 * 10¹  + 15/8 * 10⁰] =  259980

Method 2:

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First let us assume number can start with '0' also. Total no of such no : 5*4*3*2  =120.

Frequency of each digit at unit or tens or hundred or thousand place  = 120/5 = 24.

Total sum of digits at unit place  =  24[0+1+2+3+4] = 240

Total sum of all digits  = 240[

So, total sum = 24*[10³ + 10²  + 10¹  + 10⁰ ] = 266640

In this sum ,we have included those no's also which starts with '0'.so we have to deduct those sum.

 No. of no starting with '0' = 4*3*2 = 24

Hence frequency of each digit {1,2,3,4} at unit ,tens ,hundred place  = 24/4 =6 

sum of digits at unit place =  6*[1+2+3+4]= 60. 

Total sum of no starting with '0' = 60*[10²  + 10¹  + 10⁰ ] = 6660

Hence required sum is  = 266640 - 6660  = 259980

Comments

"Now expected value of first digit in any given no. = (1 +2 +3 +4)/4 =  5/2"  how do you calculate ?