Sum of all $4$ Digit no. using the digit $1,2,2,3,4,5,5$, The Sum of all $4$ Digit no. using the digit $1,2,2,3,4,5,5$, when repetition of digit is not allowed
My Try:: 
Here we are selecting $4$ Digit from a set $\{1,2,2,3,4,5,5\}$ in which we select $2$ or $5$ up to twice, each
So we will form $2$ cases::
case (I) : When $4$ Digit selected no. contain one duplicate element (Like $\{1,2,2,3,4,5\}$ or $\{1,2,3,4,5,5\}$)
case (II) : When $4$ Digit selected no. contain Two duplicate element (Like $\{2,2,3,4,5,5\}$ or $\{1,2,2,4,5,5\}$ or $\{1,2,2,3,5,5\}$)
Now I Did not understand how can i proceed after that
plz help me , Thanks

First, we approach it for the case where there are no repeated digits. We have to use $\{1, 2, 3, 4, 5\}$. There are $5*4*3*2 = 120$ such integers. The average value of each digit is $\frac{1+2+3+4+5}{5}=3$. So this yields a sum of $120*3*1111$.
Now, we approach it for the case where there are repeated digits.
Consider the case where we use 2 2's, and 2 other distinct digits. There are $4*3*{4 \choose 2} = 72$ such numbers. The average value of each digit is $\frac{1}{2} \times 2 + \frac{1}{2} \times \frac{1+3+4+5}{4} = 2.625$. Hence, the sum is $72 * 2.625 * 1111$.
Cosnider the case where we use 2 5's, and 2 other distinct digits. As before, there are 72 such numbers. The average value of each digit is $\frac{1}{2} \times 5 + \frac{1}{2} \times \frac{1+2+3+4}{4} = 3.75$. Hence the sum is $72 * 3.75 * 1111$.
Finally, add the cases where we use 2 2's and 2 5's. This is similar to the previous case and left to you.
