<s>[INST] 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.
 [/INST] 
How many six-digit numbers can be formed using numbers from the set $\{1, 2, 3, 4, 5\}$ with replacement if the digit $2$ must appear at least three times?

There are $\binom{6}{k}$ ways of choosing exactly $k$ positions for the $2$'s and $4^{6 - k}$ ways to fill the remaining $6 - k$ positions with a number different from $4$.  The number of six-digit numbers that can be formed using numbers from the set $\{1, 2, 3, 4, 5\}$ in which the digit $2$ appears at least three times can be found by adding the number of outcomes in which the digit $2$ appears exactly three times, exactly four times, exactly five times, and exactly six times
$$\sum_{k = 3}^{6} \binom{6}{k}4^{6 - k} = \binom{6}{3}4^3 + \binom{6}{4}4^2 + \binom{6}{5}4^1 + \binom{6}{6}4^0 = 1545$$
(as Dionel Jaime found) or by subtracting the number of outcomes in which the digit $2$ appears fewer than three times from the total number of words that can be formed from the five numbers in the set when those numbers are used with replacement
$$5^6 - \sum_{k = 0}^{2} \binom{6}{k}4^{6 - k} = 5^6 - \left[\binom{6}{0}4^6 + \binom{6}{1}4^5 + \binom{6}{2}4^4\right] = 1545$$

Where did you make your mistake?

By designating three positions for the $2$'s and then filling the remaining three positions with any of the five numbers in the set, you counted cases in which $2$ appears more than three times multiple times.
You counted cases in which the digit $2$ appears four times four times, once for each of the $\binom{4}{3}$ ways you could designate three of the four $2$'s as your three $2$'s.  To see this, observe that you count the number $232422$ four times:
$$\color{blue}{2}3\color{blue}{2}4\color{blue}{2}2$$
$$\color{blue}{2}3\color{blue}{2}42\color{blue}{2}$$
$$\color{blue}{2}324\color{blue}{22}$$
$$23\color{blue}{2}4\color{blue}{22}$$
You counted cases in which the digit $2$ appears five times ten times, once for each of the $\binom{5}{3}$ ways you could designate three of your five $2$'s as your three $2$'s.  You counted cases in which the digit $2$ appears six times twenty times, once for each of the $\binom{6}{3}$ ways you could designate three of your six $2$'s as your three $2$'s.  Notice that 
$$\binom{6}{3}4^3 + \binom{4}{3}\binom{6}{4}4^2 + \binom{5}{3}\binom{6}{5}4^1 + \binom{6}{3}\binom{6}{6}4^0 = 2500$$ 
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