18-04-09 Sum of all numbers

do you want jee aspirants or jee finished ones to try ????

There will be 5! times for each number in units place .
Sum of the unit = 5!(1+1+3+4+6+9) = 5!(24)

There will be 5! times each number in the tens place
Sum = 50*4!(24)

There will be 5! times each number in the hundreds place
Sum = 400*5*3!(24)

There will be 5! times each number in the thousands place
Sum = 4000*5*3*2!(24)

There will be 5! times each number in the ten thousands place
Sum = 5*4*3*2*1!(24)

There wil be 5! times each number in 1 lakhs place
Sum = 5!(24)*100000
Total
5!(24)*100000+5!(24)*10000 + 5!*1000(24) + 100*5!(24) + 10*5!(24) + 5!(24)
5!(24)[111111]

The sum of all 6 digit numbers that can be formed using

1,1, 3, 4, 6, 9 is = 5!(24)[111111]

Bhaiyya check it out

itna confident hai toh sahi hi hoga !

is repetition allowed????????????????????????????????????

Sir take any one number in the units place
Lets say we take 9
Then the remaining 5 numbers can be rearranged amongst itself in 5!
Therefore all the numbers will be repeated 5! times

Is my approach rite?

Since there are 2 '1's therefore the final answer should be divided by 2!
5!(24)[111111] /2
5!*12*111111

Sir check post #10 you will have to divide the answer by 2

ok 1 has to be considered only once

ok
Therefore the total will be
5!(23)*111111/2

ans is 222222 X 6! not as dirty as i thought it to be

fix a place
if occupied by 1 tot nos possible = 5! contributes total 5!
if occ by others (3,4,6,9) nos possible =5!/2 contributes 11X5!

tot per digit = 12X5!

so grand total is 111111 X12X5!