Six boxes are numbered from $1$ through $6$. How many are there to distribute $20$ identical balls in those boxes (some of the boxes may be empty)? Six boxes are numbered from $1$ through $6$. How many ways are there to distribute $20$ identical balls in those boxes (some of the boxes may be empty)?
Well, my solution goes like this:

Now, we can solve this by considering those $20 $ balls as objects and we can partition them in $5$ walls (say) such that each partition represents to a number of balls in one box.  So we have $25$ objects and we can arrange them $25!$ ways. But we have $20$ objects of same kind and $6$ objects of the other kind. So , the total no. of ways this can be arranged is $25!/20!5!$.

We can also solve this in the following way:

We have $20$ objects . So, we have $21$ spaces for the placement of the $5$ walls and we can place this in $21\choose 5$ ways.By doing this all the $20$ objects get partitioned among the $6$ boxes considered.

However, the answer in both the cases must be same. But why are they different? Why isn't at least one of the methods valid? Where is the problem occuring? I am not quite getting it...
 A: An alternative explanation for why the 2nd line of reasoning is false :
You don't have $21$ objects.  You have $20 + 5$ objects.  The reason is that each ball is an object, and each wall is an object.  You need $(6-1)$ walls in order to create $(6)$ regions, each of which is separated by a wall.
So, the enumeration is the number of ways of selecting $[6-1]$ objects out of $(20 + [6-1])$ objects.

In the 2nd line of reasoning, you suggest that you have $21$ objects, instead of $25$ objects.  However, the whole point of considering the selection of $n$ objects, $k$ at a time, without replacement is so that a bijection can be created between the set of all pertinent ordered $k$-tuples, and the set of all satisfying solutions to
$$x_1 + x_2 + \cdots + x_6 = 20.$$
The set of $k$-tuples in the first line of reasoning facilitates the necessary bijection.  The set of $k$-tuples in the second line of reasoning does not yield any corresponding bijection.

Edit
Reacting to the comment of Stinking Bishop following the posed question:
If the constraint is added that each of the $4$ interior variables must be $\geq 1$, the effect is that the equation requiring non-negative integer solutions is altered to
$$x_1 + x_2 + \cdots + x_6 = 16,$$
which would result in $~\displaystyle \binom{16 + [6-1]}{6-1} = \binom{21}{5}$ solutions.
