Solveeit Logo

Question

Mathematics Question on Arithmetic Progression

The sum of all the five digit numbers formed with the digits 1,2,3,4,51, 2, 3, 4, 5 taken all at a time, is

A

15(5!)15 (5!)

B

39999603999960

C

39900003990000

D

None of these

Answer

39999603999960

Explanation

Solution

Sum of unit's place =4!(1+2+3+4+5)= 4!(1 + 2 + 3 + 4 + 5) =24×15=360 = 24 \times 15 = 360 [Since 11 will come 2424 times in unit place and so the other digits if 55 digit number is formed with digits 1,2,3,4,51, 2, 3, 4, 5 taken all at a time] So the sum in tens, hundreds, thousands and ten thousands places is =360×10000+360×1000+360×100+360×10+360= 360 \times 10000 + 360 \times 1000 + 360 \times 100 + 360 \times 10 + 360 =3999960 = 3999960.