Q:

Where does the $50^*$ in $SU(5): 10\otimes10= 5^*\oplus45^*\oplus 50^*$ in A. Zee QFT?

See A. Zee, QFT in a nutshell, Appendix B, eq. (24) (p. 469 in first edition with a typo $55^*\to50^*$, cf. Zee errata; p. 530 in second edition.)
Where does the $50^*$ in $SU(5)$: $$10\otimes10= 5^*\oplus 45^*\oplus 50^*\tag{24}$$ come from? 

A:

Using the Littlewood-Richardson (LR) rules for Young tableaux, one may show that
$$ \begin{array}{c} [~~]\cr [~~] \end{array} 
\quad\otimes\quad \begin{array}{c} [a]\cr [b] \end{array}
\quad=\quad\begin{array}{c} [~~]\cr [~~] \cr [a]\cr [b]\end{array}  
\quad\oplus\quad\begin{array}{rl} [~~]&[a]\cr [~~]&\cr [b] \end{array} 
\quad\oplus\quad\begin{array}{rl} [~~]&[a]\cr [~~]&[b] \end{array} 
\quad\oplus\quad\underbrace{\begin{array}{rcl} [~~]&[a]&[b]\cr [~~] \end{array}}_{\text{forbidden}}  $$
The last extension is forbidden because of the LR rule that When reading the extension in arabic order, one must demand that the number of $a$'s $\geq$ the number of $b$'s $\geq$ the number of $c$'s$\geq\ldots $ at any point while reading, cf. Refs. 1 and 2.
For $su(N)$ irreps, the corresponding dimensions read
$$\frac{(N-1)N}{2}\otimes\frac{(N-1)N}{2}
$$
$$=\frac{(N-3)(N-2)(N-1)N}{4!}
\oplus\frac{(N-2)(N-1)N(N+1)}{2\cdot 4}
\oplus\frac{(N-1)N^2(N+1)}{2\cdot 3\cdot 2} .$$
For $N=5$, this becomes
$${\bf 10}\otimes {\bf 10} = \bar{\bf 5}
\oplus \bar{\bf 45}
\oplus \bar{\bf 50}.$$
References:

H. Georgi, Lie Algebras in Particle Physics, 1999, Section 13.2.
J.J. Sakurai, Modern Quantum Mechanics, 1994 (1st edition), Section 6.5.

