Let $V(\lambda)$ be a highest weight module of a semi-simple Lie algebra with highest weight $\lambda$. The Weyl dimension formula is $\dim V(\lambda) = \frac{\prod_{\alpha>0} (\lambda+\rho, \alpha)}{\prod_{\alpha>0}(\rho, \alpha)}$. By multiplying $\frac{2}{(\alpha, \alpha)}$ for each $\alpha$, we have $\dim V(\lambda) = \frac{\prod_{\alpha>0} \langle \lambda+\rho, \alpha \rangle}{\prod_{\alpha>0} \langle \rho, \alpha \rangle }$. Here $\langle \cdot, \cdot \rangle$ the form such that $\langle \lambda_i, \alpha_j \rangle = \delta_{ij}$ where $\lambda_j$ are fundamental weights and $\alpha_j$ are simple roots. Let $\Delta^{\vee}$ be a base of $\Phi^{\vee}$ (dual root system). Then $\alpha^{\vee} = \sum_{i} k_i \alpha_i^{\vee} $. How to compute the coefficients $k_i$? If we can compute $k_i$, then we know how to use the Weyl dimension formula.
For type $G_2$, I computed that $\alpha_1=2\lambda_1-\lambda_2$, $\alpha_2=-3\lambda_1+2\lambda_2$, $\alpha_1+\alpha_2=-\lambda_1+\lambda_2$, $2\alpha_1+\alpha_2=\lambda_1$, $3\alpha_1+\alpha_2=3\lambda_1-\lambda_2$, $3\alpha_1+2\alpha_2=\lambda_2$. These are all positive roots. $\rho=\lambda_1+\lambda_2$.
Let $\lambda=m_1\lambda_1+m_2\lambda_2$. How can we obtain the formula $\dim V(\lambda) = 1/120 (m_1+1)(m_2+1)(m_1+m_2+2)(m_1+2m_2+3)(m_1+3m_2+4)(2m_1+3m_2+5)?$