Prove that each number $a$ in a Pascal's triangle decreased by $1$ is equal to the sum of the numbers within the parallelogram bounded by the sides of Prove that each number $a$ in a Pascal's triangle decreased by $1$ is equal to the sum of the numbers within the parallelogram bounded by the sides of the triangle and the diagonalsgoing through $a$ (The numbers in the diagonals are not included )
The above figure shows the representation in a Pascal's triangle as an example .
In order to solve the above problem I proved that: Each number $a$ in Pascal's triangle is equal to the sum of the numbers in the previous right diagonal,  starting from its leftmost number through the number through the number which is located in the left diagonal as $a$.
The above figure represents an example of such an arrangement .
Also I proved the same for left diagonals as well,
Each number $a$ in Pascal's triangle is equal to the sum of the numbers in the previous left diagonal,  starting from its rightmost number through the number through the number which is located in the right  diagonal as $a$.

Now, how to apply this and then prove the original result?

Here is a sketch of a proof by induction of your claim. I illustrate the example of $\color{blue}{\binom{7}{3}}=35$.
By the hockey stick identity, you know that
$$
\color{blue}{\binom{7}3}=\color{purple}{\binom{6}3}+\color{red}{\text{sum of red numbers}}
$$
If we apply the inductive hypothesis to $\color{purple}{\binom{6}3}$, we get that
$$
\color{purple}{\binom{6}3}=\text{sum of black numbers}
$$
Combining these,
$$
\color{blue}{\binom{7}3}=\color{red}{\text{sum of red numbers}}+\text{sum of black numbers}
$$
Since the red and black numbers combine to form the required parallelogram for $\color{blue}{\binom 73}$, the proof is complete.

