Key Concepts and Formulas

• Combinations: The number of ways to choose $r$ objects from a set of $n$ distinct objects is given by ${}^nC_r = \frac{n!}{r!(n-r)!}$.
• Pascal's Identity: ${}^nC_r + {}^nC_{r-1} = {}^{n+1}C_r$
• Hockey Stick Identity: ${}^rC_r + {}^{r+1}C_r + ... + {}^nC_r = {}^{n+1}C_{r+1}$

Step-by-Step Solution

Step 1: Expand the summation. What we are doing: We are writing out the summation explicitly to better visualize the terms. $\sum_{k=0}^6 {}^{51-k}C_3 = {}^{51}C_3 + {}^{50}C_3 + {}^{49}C_3 + {}^{48}C_3 + {}^{47}C_3 + {}^{46}C_3 + {}^{45}C_3$ Reasoning: This is a direct application of the summation notation.

Step 2: Rearrange the terms in ascending order of the lower index. What we are doing: We are simply reordering the terms to prepare for applying the Hockey Stick Identity. ${}^{45}C_3 + {}^{46}C_3 + {}^{47}C_3 + {}^{48}C_3 + {}^{49}C_3 + {}^{50}C_3 + {}^{51}C_3$ Reasoning: Reordering the terms doesn't change the sum's value.

Step 3: Add and subtract ${}^{45}C_4$. What we are doing: We add and subtract the term ${}^{45}C_4$ to manipulate the expression so that we can use Pascal's Identity. ${}^{45}C_3 + {}^{46}C_3 + {}^{47}C_3 + {}^{48}C_3 + {}^{49}C_3 + {}^{50}C_3 + {}^{51}C_3 = ({}^{45}C_4 + {}^{45}C_3) + {}^{46}C_3 + {}^{47}C_3 + {}^{48}C_3 + {}^{49}C_3 + {}^{50}C_3 + {}^{51}C_3 - {}^{45}C_4$ Reasoning: Adding and subtracting the same term doesn't change the value of the expression.

Step 4: Apply Pascal's Identity repeatedly. What we are doing: We are using the identity ${}^nC_r + {}^nC_{r-1} = {}^{n+1}C_r$ to simplify the expression. $({}^{45}C_4 + {}^{45}C_3) + {}^{46}C_3 + {}^{47}C_3 + {}^{48}C_3 + {}^{49}C_3 + {}^{50}C_3 + {}^{51}C_3 - {}^{45}C_4 = {}^{46}C_4 + {}^{46}C_3 + {}^{47}C_3 + {}^{48}C_3 + {}^{49}C_3 + {}^{50}C_3 + {}^{51}C_3 - {}^{45}C_4$ $= {}^{47}C_4 + {}^{47}C_3 + {}^{48}C_3 + {}^{49}C_3 + {}^{50}C_3 + {}^{51}C_3 - {}^{45}C_4$ $= {}^{48}C_4 + {}^{48}C_3 + {}^{49}C_3 + {}^{50}C_3 + {}^{51}C_3 - {}^{45}C_4$ $= {}^{49}C_4 + {}^{49}C_3 + {}^{50}C_3 + {}^{51}C_3 - {}^{45}C_4$ $= {}^{50}C_4 + {}^{50}C_3 + {}^{51}C_3 - {}^{45}C_4$ $= {}^{51}C_4 + {}^{51}C_3 - {}^{45}C_4$ $= {}^{52}C_4 - {}^{45}C_4$ Reasoning: We repeatedly apply Pascal's Identity, combining adjacent terms until we arrive at a simplified expression.

Step 5: Final Result What we are doing: State the final simplified form of the summation. $\sum_{k=0}^6 {}^{51-k}C_3 = {}^{52}C_4 - {}^{45}C_4$ Reasoning: This is the final result after repeated application of Pascal's identity.

Common Mistakes & Tips

• Forgetting Pascal's Identity: Make sure to memorize Pascal's Identity, as it's fundamental to solving many combination problems.
• Not recognizing the Hockey Stick Identity: The Hockey Stick Identity can provide a shortcut but is not strictly needed here.
• Sign Errors: Be very careful with the signs when adding and subtracting terms.

Summary

We were asked to evaluate the summation $\sum_{k=0}^6 {}^{51-k}C_3$. We first expanded the summation and then rearranged the terms. We then added and subtracted ${}^{45}C_4$, which allowed us to repeatedly apply Pascal's Identity. This simplified the expression to ${}^{52}C_4 - {}^{45}C_4$.

Final Answer

The final answer is \boxed{{}^{52}{C_4} - {}^{45}{C_4}}, which corresponds to option (D).