9do you want jee aspirants or jee finished ones to try ????
11There 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]
1itna confident hai toh sahi hi hoga !
62well virang.. I think there is a small mistake there..
will the unit digit be repeated that many no of times for every digit>?
24is repetition allowed????????????????????????????????????
62
Lokesh Verma
·2009-04-18 00:06:59
repitition is allowed exactly as many times as the numbers appear
11
virang1 Jhaveri
·2009-04-18 00:07:49
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?
11
virang1 Jhaveri
·2009-04-18 00:11:17
Since there are 2 '1's therefore the final answer should be divided by 2!
5!(24)[111111] /2
5!*12*111111
62
Lokesh Verma
·2009-04-18 00:11:51
virang but here there are 2 one's
so that will cause a trouble..
11
virang1 Jhaveri
·2009-04-18 00:14:19
Sir check post #10 you will have to divide the answer by 2
62
Lokesh Verma
·2009-04-18 00:29:23
virang .. i think the treatment will be different for digits 1 and the other digits..
Think once..
11
virang1 Jhaveri
·2009-04-18 00:43:24
ok 1 has to be considered only once
ok
Therefore the total will be
5!(23)*111111/2
9
Celestine preetham
·2009-04-18 01:03:35
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!
62
Lokesh Verma
·2009-04-18 20:41:10
Good work celestine..
It would have become dirtier if there was a zero.. that is why i did not include a zero in the question..