Evaluate without a calculator:

$$\mathbf{12\binom{3}{3} + 11\binom{4}{3} + 10\binom{5}{3} + \cdots + 2\binom{13}{3} +\binom{14}{3} = \;?}$$

$$\small{\text{$ \begin{array}{l} \mathbf{ 12\binom{3}{3} + 11\binom{4}{3} + 10\binom{5}{3} + 9\binom{6}{3} + 8\binom{7}{3} + 7\binom{8}{3} + 6\binom{9}{3} + 5\binom{10}{3} + 4\binom{11}{3} + 3\binom{12}{3} + 2\binom{13}{3} + 1\binom{14}{3} } \\ \\ = \mathbf{ \binom{4}{4} + \binom{5}{4} + \binom{6}{4} + \binom{7}{4} + \binom{8}{4} + \binom{9}{4} + \binom{10}{4} + \binom{11}{4} + \binom{12}{4} + \binom{13}{4} + \binom{14}{4} + \binom{15}{4} }\\\\ = \mathbf{ \binom{16}{5} }\\\\ = \mathbf{ \dfrac{16!}{5!\cdot 11!} }\\\\ = \mathbf{ \dfrac{12\cdot 13\cdot 14\cdot 15\cdot 16}{5!} }\\\\ = \mathbf{ \dfrac{12\cdot 13\cdot 14\cdot 15\cdot 16}{1\cdot 2\cdot 3\cdot 4\cdot 5} }\\\\ = \mathbf{ \left(\dfrac{12}{3}\right)\cdot 13 \cdot \left(\dfrac{14}{2}\right)\cdot \left(\dfrac{15}{5}\right)\cdot \left(\dfrac{16}{4}\right) }\\\\ = \mathbf{4 \cdot 13 \cdot 7 \cdot 3 \cdot 4}\\\\ = \mathbf{13 \cdot 16 \cdot 21}\\\\ = \mathbf{4368} \end{array} $}}$$

see Hockey Stick Pattern: http://ptri1.tripod.com/

If a diagonal of numbers of any length is selected starting at any of the 1's bordering the sides of the triangle and ending on any number inside the triangle on that diagonal, the sum of the numbers inside the selection is equal to the number below the end of the selection that is not on the same diagonal itself. If you don't understand that, look at the drawing. 1+6+21+56 = 84 1+7+28+84+210+462+924 = 1716 1+12 = 13

Thanks Heureka,

That web site that you have referenced looks very interesting.

I know that there are a lot of features on Pascal's triangle that I am not aware of or that I do not retain in my memory.  So thank you.  I think I will store that site away in our reference material :)