Question

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Annotation
1.
Find the sum of all five digited numbers that can be formed using the digits 1, 2, 3, 7, 9.
digits
5​

Answer 1

If 5 is at the unit's place, the remaining digits 1,2,3 and 4 can be arranged among themselves in 4! ways. Hence, there are exactly 4! five-digit numbers using the given digits.

Similarly, there are exactly 4! five-digit numbers using the given digits. Similarly for other digits.

Hence the total sum of digits at unit's place for all numbers possible is 4!(5+4+3+2+1)

Similarly, the total sum of digits at ten's place for all numbers possible is 10×4!(5+4+3+2+1)

Similarly, for hundred's, thousand's and ten thousand's place the sum are 10

2

×4!(5+4+3+2+1),10

3

×4!(5+4+3+2+1),10

4

×4!(5+4+3+2+1) respectively.

Hence, the required sum

=(4)!(1+2+3+4+5)(1+10+10

2

+10

3

+10

4

)

=24×15×(

10−1

10

5

−1

)

=24×15×11111

=3999960