Calculate the value of the sum: $$$ \begin{pmatrix} 4 \\ 0 \end{pmatrix}+ \begin{pmatrix} 4 \\ 1 \end{pmatrix}+ \begin{pmatrix} 4 \\ 2 \end{pmatrix}+ \begin{pmatrix} 4 \\ 3 \end{pmatrix}$$$
See development and solution
Development:
Applying the formulas we have:
$$$ \begin{array}{rl} \begin{pmatrix} 4 \\ 0 \end{pmatrix}+ \begin{pmatrix} 4 \\ 1 \end{pmatrix}+ \begin{pmatrix} 4 \\ 2 \end{pmatrix}+ \begin{pmatrix} 4 \\ 3 \end{pmatrix} &= 1+4+\dfrac{4!}{2!2!}+4+1 \\ &=1+4+6+4+1=16 \end{array}$$$ note the symmetry of the numbers.
Solution:
$$16$$