Do it here

I'll provide a $\color{#3D99F6}{\text{general formula}}$ for this problem statement:

The number of distinct terms in expansion of $( a_{\color{#D61F06}1} + a_{\color{#D61F06}2} + \ldots + a_\color{#D61F06}n)^{\color{#69047E}k} = \dbinom{\color{#69047E}{k} + \color{#D61F06}n - 1 }{\color{#D61F06}{ n} - 1}$

Therefore, the number of distinct terms in $( a + b + c)^{20}$ is $\dbinom{20 + 3 - 1}{3 - 1 } = 231$

General formula to get number of terms in multinomial expansion $(a_1+ a_2+a_3+...a_k)^n$ is given by ${n+k-1 \choose k-1}$

So now it is easier to get number of terms of equation $(a+b+c)^{20}$ which is ${20+3-1 \choose 3-1}$ whose value is $\boxed{231}$

By stars and Bars

Our approach is $_3 H_{20}$ = $_{3 + 20 - 1} C_{20}$ = $_{22} C_{2}$ = 231

Answer: $\boxed{231}$