A probability problem by Nivedit Jain
Probability
Level
3
What is the sum of all 4 digit numbers that can be formed with digits 0,1,2, and 3 if repetition is not allowed?
The answer is 38664.
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1 solution
Use the formula sum of numbers is (n-1)!(sum of digits)(10^9 -1)/9 For 4 digits we have Sum =3! 6 1111 Now we have also included numbers starting with 0 so subtracting them Sum of 3 digit no = 2! 6 111 Final answer is 1 -2 We get 6[6666 - 222] = 38664