How many $6$ digit numbers can you make with the numbers $\{1, 2, 3, 4, 5\}$ so that the digit $2$ appears at least 3 times? I can't seem to understand.  

I thought about it this way. We take a first example:  

$222aaa$ for each blank space we have $5$ positions, so in this case the answer would be $125$ ($5 \times 5 \times 5$) numbers.

Now the position of the twos can change so we calculate the numbers for that taking this example:  

$222aaa$ and the possibilities for this are the number of ways you can arrange $6$ digits ($6!$) and divide by repetitions so divided by $2 \cdot (3!)$ so answer $= 20$

So final answer should be $125 \cdot 20 = 2500$. But this answer is wrong and I don't understand why.
Let ${S_n}$ be the number of ways you can have the  digit 2 appears ONLY 3, 4 ,5 , and 6 times where $ n = 3 , 4 , 5, 6$.  We can then add $S_3 + S_4 + S_5 + S_6$ to get our answer.  To find $S_n$, note that there are ${6}\choose {n}$ ways of arranging the 2's, and for each of those there are $6-n$ places for the other digits; however, we will not include 2 in these other digits or else we will end up overcounting.  So we $4^{(6-n)}$ different numbers we can make.

So $S_3 =$ ${6}\choose{3} $$4^3$ 

$S_4 =$ ${6}\choose {4}$$4^2$

$S_5 =$ ${6}\choose{5} $$4 = 24$

$S_6 =$ ${6}\choose{6}$ $=1$

Now just add these numbers up.
Since you have only $5$ numbers, you can try it making cases


  Case $1.$ When there are $3~2's$


$3$ spaces left


  Case $1.a)$  When all the three numbers are different 


Total numbers= $4\times 3\times 2=24$

Total permutation=$\frac{6!}{3!}=120$

Total cases=$24\times 120=2880$


  Case $1.b)$ When two of three numbers are same


Total numbers=$12$

Total permutation=${6!}{3!2!}=60$

Total cases=$12\times 60=720$


  Case 2. When there are $4 ~2's$


$2$ places left


  Case $2.a)$ When the two numbers are different


Total numbers=$4\times 3=12$

Total permutation=$\frac{6!}{4!}$

Total cases=$12\times 30=360$


  Case $2.b) When the two numbers are same


Total numbers=$4$

Total permutation=$\frac{6!}{4!2!}=15$

Total cases=$4\times 15=60$


  Case $3.$ When there are five $2's$


Total cases=$4\times 6=24$


  Case $4.$ When there are  six $2's$


Total cases=$1$

Adding all will give $2880+720+360+60+24+1=4045$