Problems from Identity of symmetry

Calculate the value of the sum: $(\begin{matrix} 4 \\ 0 \end{matrix}) + (\begin{matrix} 4 \\ 1 \end{matrix}) + (\begin{matrix} 4 \\ 2 \end{matrix}) + (\begin{matrix} 4 \\ 3 \end{matrix})$

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Development:

Applying the formulas we have:

$\begin{aligned} (\begin{array}{c} 4 \\ 0 \end{array}) + (\begin{array}{c} 4 \\ 1 \end{array}) + (\begin{array}{c} 4 \\ 2 \end{array}) + (\begin{array}{c} 4 \\ 3 \end{array}) & = 1 + 4 + \frac{4!}{2! 2!} + 4 + 1 \\ = 1 + 4 + 6 + 4 + 1 = 16 \end{aligned}$ note the symmetry of the numbers.

Solution:

$16$

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