Pela propriedade da soma na coluna do triângulo de Pascal, temos que:
$\,\binom{2}{2}\,+\,\binom{3}{2}\,+\,\binom{4}{2}\,=\,\binom{5}{3}\,=\,10$
Pela propriedade da soma na diagonal do triângulo de Pascal (veja a figura), temos que:
$\,\binom{2}{0}\,+\,\binom{3}{1}\,+\,\binom{4}{2}\,+\,\binom{5}{3}\,=\,\binom{6}{3}\,=\,20$
Então:
$\,\frac{10 \, \left[\large {\binom{3}{0}\,+\,\binom{3}{1}\,+\,\binom{3}{2}\,+\,\binom{3}{3}} \right]\,+\,2\,\left[\,\large {\binom{2}{2}\,+\,\binom{3}{2}\,+\,\binom{4}{2}} \right]\phantom{XX}}{ {\large \binom{2}{0}\,+\,\binom{3}{1}\,+\,\binom{4}{2}\,+\,\binom{5}{3}}}\,=\,$ $\,\dfrac{10\centerdot 8 \,+\, 2\centerdot 10}{20}\,=\,\dfrac{100}{20}\,=\,5$