If we look at the number $a_6a_5a_4a_3a_2a_1a_0$:
$$a_6a_5a_4a_3a_2a_1a_0 = a_610^6+a_510^5+ \dots +a_010^0 \\ \equiv a_6(-1)^6 + \dots+a_0(-1)^0 \mod 11 = a_6 -a_5+a_4-a_3+a_2-a_1+a_0 \mod 11$$
So if the number will be divisible by 11 we require
$$a_6 -a_5+a_4-a_3+a_2-a_1+a_0 = 11m$$
We take note that $0 \leq a_i \leq 9$ so possible values of $11m$ are limited to:
$$-27 = 4\cdot0-3\cdot9\leq 11m \leq 4\cdot 9-3\cdot0 = 36$$
Which really means that $11m \in \{-22,-11,0,11,22,33\}$
The other requirement of the question was that $a_6 + \dots +a_0 = 59$. From this and the above equation (not sure how to number them and align them nicely in TeX) we add and subtract and get much nicer equations:
$$a_6 +a_4+a_2+a_0 = \frac{59+11m} 2
\\a_5+a_3+a_1 =\frac{ 59-11m}2$$
Of course the LHS is whole, so the right hand side must be as well, which means $m$ needs to be odd. So we reduce our options to:
$$11m \in \{-11,11,33\} \implies \frac{59+11m} 2 \in \{24,35,46\}$$
Of course the sum of four digits can't be $46$ from our above inequality on $a_i$, and similarly
$$\frac{59-11m} 2 \in \{35,24\}$$ But the sum of three digits can't be $35$, so we're left with
$$a_6 +a_4+a_2+a_0 = 35
\\a_5+a_3+a_1 =24$$
It's easy to see that the only options for the four digits is a permutation of $9998$, and the three digits must be a permutation of one of $\{699,789,888\}$.
Order doesn't matter, so basic combinatorics gives $4\cdot(3 + 3! + 1) = 40$ such numbers.
I will use the fundamental principle of counting to solve this question.
Given the set of numbers, $D = \{2, 2, 3, 3, 3, 4, 4, 4, 4\}$.
Here, count(2's) = 2, count(3's) = 3, count(4's) = 4.
We are required to construct 4-digit numbers greater than 3000. For easy visualisation, we will use dashes on paper, _ _ _ _.
Case 1: Thousandth's place is 3.
Set, $D' = \{2, 2, 3, 3, 4, 4, 4, 4\}$.
Rest of the 3 places can be filled in $3 \times 3 \times 3$ ways provided we have 3 counts of the three unique digits, for filling each of the 3 places. But count(2's) = 2, count(3's) = 2 in the new set. Deficiency for each digit is 1. (Impossible numbers - 3222, 3333).
Therefore, ways to fill = $3 \times 3 \times 3 - (1+1) = 25$.
Case 2: Thousandth's place is 4.
Set, $D' = \{2, 2, 3, 3, 3, 4, 4, 4\}$.
Similarly, the deficiency is of only digit 2, which is 1 count. (Impossible number - 4222)
Therefore ways to fill = $3 \times 3 \times 3 - 1 = 26$.
Summing each case up, we get $26 + 25 = 51$.
Hope this helps.
Best Answer
I think considering the two different cases you mentioned separately is best. To avoid further case division, I'd proceed like this:
Case with leading digit 4: since an even digit has to go in the rightmost position, there are $5 \choose 3$ ways to choose the positions of the 3 odd numbers amongst the other 5 positions, then 3 ways to order them, and $3!$ ways to order the even numbers, for 180 total.
Case with leading digit 5: similarly, there are $5 \choose 2$ ways to choose the positions of the two odd digits (both 3's), only 1 way to order them, then $\frac{4!}{2}$ ways to order the even digits (dividing by 2 since there are 2 indistinguishable 4's), for 120 total.
So all up there are 300 such numbers.