when we need to find the sum of digits such as in this case we can either form cases because the number of digits are less.
else the concept to find the sum is
ABC.
when C is 3,4,5 so AB can be arranged in 2! ways.
the sum of digits is 3+4+5 and since it is in the units digits it takes a place value of 1.
so it becomes (3+4+5)*1*2!
similarly in 10s digit, we fix 3,4,5 and the rest of the digits can be arranged in 2! ways.
so the place value is 10 and sum becomes (3+4+5)*10*2!
similarly for hundreds place it becomes (3+4+5)*100*2!
to find the sum of all digits we add these three,
it becomes 100(3+4+5)2! + 10(3+4+5)2! + 1(3+4+5)2!
taking 2! and sum of digits as common.
2!(3+4+5)*111
we can also generalise a formula from this,
3,4,5 ----> no of digits(n) = 3
the formula becomes
(n-1)! (111....n times) ( sum of digits)
so here n = 3.
putting the values in the formula,
(3-1)! (111) (3+4+5).
This gives us the answer as 2664.
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