First let me simplify the

First let me simplify the question a bit
How many distinct 4 digit numbers can be made of the numbers 1143 with out repetition.

There are many ways of approaching such questions but the simplest is to use the following formula:

"The number of permutations of n elements with n1 of repeated element, n2 of another repeated element, etc. taken n at a time. "

P=n!/(n1!)(n2!) ( please check http://cat4mba.com/math-e-book/permutations for the details)

So here n=4 i.e four different numbers and 1 is repeated twice so n1=2
Thus P=4!/2! = 12

So 12 different numbers can be made.

( Just to verify I am writing these numbers (please don’t do so in any exam)
1143, 1134, 3114, 4113, 3411, 4311, 1413, 1314, 1341, 1431, 3141, 4131)

Now let’s find out what is the sum of all the possible numbers

We have 12 different numbers i.e 12 numbers with a uint place, 10th place, 100th place and 1000th place.

In all these places the digits will be distributed proportionally i.e
In unit place we will have 6 1’s , 3 3’s and 3 4’s

So the sum is 6*1 + 3*3 + 3 * 4 = 6 + 9 + 12 =27

Thus Sum of all the places 27 + 27*10 + 27*100 + 27 * 1000 = 27 * 1111 = 29997.

A simple formula to get this

A simple formula to get this solution can be

(n-1)*(sum of digits)*(1111....n times)

so ans = (4-1)*9*(1111) = 29997

@Divya

Divya this is not d formula....
If n are all distinct integers.
Dn sum:(n-1)!*(Sum of d digits)*(1111...upto n digits)
As here 1 is repeted for twice

So formula wud be
(n-1)!/2!*(Sum of d digits)*(1111..uto n digits)[As all no.s will b distinct]
=3!/2!*(1+1+4+3)*(1111)
=27*1111=29997

Giv ur best to d world.
Nd d best will cm back to u......

Regards,
Dipanjan......