What is the exact numerical value of the combination sum $\left(\genfrac{}{}{0ex}{}{50}{4}\right)+{\sum }_{r=1}^6\left(\genfrac{}{}{0ex}{}{56-r}{3}\right)$?

Step 1 (Expand the summation): We write out the terms in the summation ${\sum }_{r=1}^6\left(\genfrac{}{}{0ex}{}{56-r}{3}\right)$ progressively:

• For $r=1$: $\left(\genfrac{}{}{0ex}{}{55}{3}\right)$

• For $r=2$: $\left(\genfrac{}{}{0ex}{}{54}{3}\right)$

• For $r=3$: $\left(\genfrac{}{}{0ex}{}{53}{3}\right)$

• For $r=4$: $\left(\genfrac{}{}{0ex}{}{52}{3}\right)$

• For $r=5$: $\left(\genfrac{}{}{0ex}{}{51}{3}\right)$

• For $r=6$: $\binom{50}{3}

So the total expression is:

Step 2 (Apply Pascal's Identity): Pascal's Identity states that:

Let's simplify our expression from left to right:

1. $\left(\genfrac{}{}{0ex}{}{50}{4}\right)+\left(\genfrac{}{}{0ex}{}{50}{3}\right)=\left(\genfrac{}{}{0ex}{}{51}{4}\right)$

2. Substitute this back: $S=\left(\genfrac{}{}{0ex}{}{51}{4}\right)+\left(\genfrac{}{}{0ex}{}{51}{3}\right)+\left(\genfrac{}{}{0ex}{}{52}{3}\right)+\dots$

3. $\left(\genfrac{}{}{0ex}{}{51}{4}\right)+\left(\genfrac{}{}{0ex}{}{51}{3}\right)=\left(\genfrac{}{}{0ex}{}{52}{4}\right)$

4. Continuing this inductive chain:

• $\left(\genfrac{}{}{0ex}{}{52}{4}\right)+\left(\genfrac{}{}{0ex}{}{52}{3}\right)=\left(\genfrac{}{}{0ex}{}{53}{4}\right)$

• $\left(\genfrac{}{}{0ex}{}{53}{4}\right)+\left(\genfrac{}{}{0ex}{}{53}{3}\right)=\left(\genfrac{}{}{0ex}{}{54}{4}\right)$

• ...

• Finally, we get $S=\left(\genfrac{}{}{0ex}{}{55}{4}\right)+\left(\genfrac{}{}{0ex}{}{55}{3}\right)=\left(\genfrac{}{}{0ex}{}{56}{4}\right)$

Step 3 (Calculate the combination value): Evaluate $\left(\genfrac{}{}{0ex}{}{56}{4}\right)$:

Step 4 (Conclusion): The value of the sum is exactly 367290.