How many ways can 200 identical balls be distributed into 40 distinct jars? 
How many ways can $200$ identical balls be distributed into $40$ distinct jars so that number Balls in the first $20$ jars is greater than the number of balls in the last $20$ jars?

I came up with two different solutions to the problem but the answers come out differently. The first solution uses symmetry:
There are $\binom{239}{40}$ possible ways to place $200$ balls in $40$ jars. We then subtract all possibilities where the number of balls in the first $20$ jars is equal to the number of balls in the last $20$ jars, which is equal to $\binom{119}{20}^2$. We then exploit symmetry (the number of possibilities that there are more balls in the first $20$ jars is equal to the number of possibilities that there are more balls in the last twenty jars). The final answer is 

$$\frac{\binom{239}{40} - \binom{119}{20}^2}{2}$$

The second solution uses a sum. It comes out to 

$$\sum_{k=101}^{200} \binom{k+19}{20}\binom{219-k}{20}$$

 A: An expression for the number found by means of stars and bars is:
$$\sum_{k=0}^{99}\binom{k+19}{19}\binom{200-k+19}{19}$$
We can rewrite this as:$$\sum_{i+j=238\wedge i\leq118}\binom{i}{19}\binom{j}{19}$$under the convention that $\binom{n}{m}=0$ if $m\notin\{0,1,\dots,n\}$.
Further we have:$$\binom{239}{39}=\sum_{i+j=238}\binom{i}{19}\binom{j}{19}=$$$$\sum_{i+j=238\wedge i\leq118}\binom{i}{19}\binom{j}{19}+\sum_{i+j=238\wedge j\leq118}\binom{i}{19}\binom{j}{19}+\binom{119}{19}^2$$where the first equality can be recognized as the hockey-stick equality.
This with:$$\sum_{i+j=238\wedge i\leq118}\binom{i}{19}\binom{j}{19}=\sum_{i+j=238\wedge j\leq118}\binom{i}{19}\binom{j}{19}$$
so that:$$\sum_{i+j=238\wedge i\leq118}\binom{i}{19}\binom{j}{19}=\frac{1}{2}\left[\binom{239}{39}-\binom{119}{19}^2\right]$$
A: You made an error in applying "stars and bars". You should replace $\binom{239}{40}$ with $\binom{239}{39}$,  and $\binom{119}{20}$ with $\binom{119}{19}$ and so on.
With correct expressions you obtain:
$$
\sum_{k=101}^{200} \binom{k+19}{19}\binom{219-k}{19}=\frac{\binom{239}{39} - \binom{119}{19}^2}{2}.
$$
No contradiction appears.
