Combinatorics and probability question 
Suppose we have $20$ balls numbered $1,2,...,20$. Each day we pick a single ball randomly.
1) What's the probability of picking all $20$ balls in $22$ days ?
2) What's the probability of picking only $1 \,and\, 2$ in $4$ days (Atleast 1 times each) ?

Attempts and Thoughts - 
1) I thought that we could pick $20$ days out of the $22$ - $\binom{22}{20}$. Now there are $20!$ distinct orders for all the $20$ balls in the $20$ picked days. What's left is to put $2$ balls in the days left - $20^2$ options for that and over all - $\binom{22}{20}\cdot 20!\cdot 20^2$. I thought this might be true until i encountered another approach suggesting - $\binom{20}{1}\cdot \binom{22}{3}\cdot 19! + \binom{20}{2}\cdot \binom{22}{2}\cdot \binom{20}{2}\cdot 18!$.  The leftmost part reffering to when a ball is being picked $3$ times and the right most to $2$ balls being picked twice.
Which approach is right? What have i done wrong?
2) First approach - there are exactly $2^4$ options considering $1,1,1,1$ or $2,2,2,2$. So we get - $2^4 - 2= 14$ options.
Another approach suggested $22^4 - 2\cdot 21^4 +20^4$ which comes out of basic inclusion-exclusion.
Would like to hear you thoughts about my problem. Thanks in advance.

Problem 1
Your approach to the first question is incorrect : there is overcounting happening.
To see why this is the case, let us see the steps you are using :


*

*Fix a subset of $1,...,22$ : can be done in $\binom{22}{20}$ ways.

*Arrange for $20$ balls to be picked in these turns : done in $20!$ ways.

*Two turns are left, and they may be allotted freely , hence in $20^2$ ways.
The overcounting comes from the fact that it is possible to start from different subsets and still get the same order using the free choice of the third step.

Example :
Start with the subset $\{1,...,20\}$, and arrange for ball $1$ to be picked on the first day, ball $2$ on the second day , ball $3$ on the third day etc. by the second step. Now for $21$ and $22$,  let balls $19$ be picked on day $21$ and ball $20$ on day $22$. Thus, the final order is $1,2,3,...,20,19,20$.
Now, start with the subset $\{1,...,18,21,22\}$. Arrange for ball $1$ to be picked on the first day, ball $2$ on the second day , ..., ball $18$ to be picked on the $18$th day, ball $19$ on the $21$st day and ball $20$ on the $22$nd day. Now, on the remaining days, let ball $19$ be picked on day $19$, and ball $20$ on day $20$. This gives the same order as the previous order, but with a different subset.
Therefore, a direct multiplication of the combinations resulting from the above steps is over counting.

The way the other solution takes care of this problem, is not to isolate which subset of $20$ elements contains $20$ different balls, but rather which ball appears twice/thrice (no ball appears more than thrice : this would leave $19$ different numbered balls to be distributed in $18$ days,  which  can't happen). If a number appears thrice, then all others appear exactly once : this gives one case. If one number appears twice, so must exactly one other number : this gives another case. But one of the above must happen : this gives the two cases of the second (and correct) solution.

Problem 2
Your approach is correct. The other approach seems to be of a different question : counting the ways in which at least one and at least one two occurs.  
