Permutation and CombinationHard

Question

Sum of all the numbers that can be formed using all the digits 2, 3, 3, 4, 4, 4, is:

Options

A.22222200
B.11111100
C.55555500
D.20333280

Solution

= = 10
= = 20
= = 30
Hence sum of unit places is
2 × 10 + 3 × 20 + 4 × 30 = 200
Hence required sum is
=  200 × (105 + 104 + 103 + 102 + 101 + 100)
=  200 × (111111) = 22222200

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