Key Concept: Pascal's Identity and the Hockey-stick Identity

This problem primarily relies on two fundamental identities in combinatorics:

1. Pascal's Identity: ${{}^n{C_r} + {}^n{C_{r - 1}} = {}^{n + 1}{C_r}}$ This identity states that the sum of two adjacent binomial coefficients in Pascal's triangle is equal to the coefficient directly below them. It can also be written as ${{}^n{C_r} + {}^n{C_{r + 1}} = {}^{n + 1}{C_{r + 1}}}$. We will use this form extensively.

2. Hockey-stick Identity (or Christmas Stocking Identity): ${{}^k{C_k} + {}^{k + 1}{C_k} + {}^{k + 2}{C_k} + \dots + {}^{n}{C_k} = {}^{n + 1}{C_{k + 1}}}$ This identity represents the sum of binomial coefficients along a diagonal in Pascal's triangle. If the sum does not start from ${^k{C_k}}$, it can be modified as: $\sum_{i=r}^{n} {^{i}{C_k}} = {^{n+1}{C_{k+1}}} - {^{r}{C_{k+1}}}$

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Step-by-Step Derivation

The problem asks for the value of the summation: $S = \sum\limits_{r = 0}^{20} {{}^{50 - r}{C_6}}$

1. Expand the Summation: First, let's write out the terms of the summation to understand the pattern. For $r=0$, the term is ${}^{50 - 0}{C_6} = {}^{50}{C_6}$. For $r=1$, the term is ${}^{50 - 1}{C_6} = {}^{49}{C_6}$. ... For $r=20$, the term is ${}^{50 - 20}{C_6} = {}^{30}{C_6}$.

So, the summation can be written as: $S = {}^{50}{C_6} + {}^{49}{C_6} + {}^{48}{C_6} + \dots + {}^{31}{C_6} + {}^{30}{C_6}$ For easier application of the identities, it's often helpful to write the sum in ascending order of the upper index: $S = {}^{30}{C_6} + {}^{31}{C_6} + \dots + {}^{49}{C_6} + {}^{50}{C_6}$

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Method 1: Iterative Application of Pascal's Identity

This method involves skillfully introducing and subtracting a term to initiate a chain reaction using Pascal's Identity ${{}^n{C_r} + {}^n{C_{r + 1}} = {}^{n + 1}{C_{r + 1}}}$.

2. Introduce a Catalyst Term: To start combining terms using Pascal's Identity, we need a ${^n{C_{r+1}}}$ term corresponding to the initial ${^n{C_r}}$ term in our sum. Our first term is ${^{30}{C_6}}$. If we had a ${^{30}{C_7}}$ term, we could combine them. So, we add and subtract ${^{30}{C_7}}$. $S = \left( {}^{30}{C_6} + {}^{30}{C_7} \right) + {}^{31}{C_6} + \dots + {}^{50}{C_6} - {}^{30}{C_7}$ Explanation: We add ${^{30}{C_7}}$ to the sum to create the necessary condition for Pascal's Identity. To keep the sum's value unchanged, we immediately subtract the same term.

3. Apply Pascal's Identity Iteratively: Now, we can apply Pascal's Identity to the first group of terms: Using ${{}^n{C_r} + {}^n{C_{r + 1}} = {}^{n + 1}{C_{r + 1}}}$ with $n=30$, $r=6$: ${}^{30}{C_6} + {^{30}{C_7}} = {^{31}{C_7}}$ Substitute this back into our expression for $S$: $S = {}^{31}{C_7} + {}^{31}{C_6} + {}^{32}{C_6} + \dots + {}^{50}{C_6} - {}^{30}{C_7}$ Next, we combine ${^{31}{C_7}}$ with the next term in the sum, ${^{31}{C_6}}$ (reordering for clarity): ${^{31}{C_6}} + {^{31}{C_7}} = {^{32}{C_7}}$ Substitute again: $S = {}^{32}{C_7} + {}^{32}{C_6} + \dots + {}^{50}{C_6} - {}^{30}{C_7}$ This pattern continues. Each time, the result of the previous application of Pascal's Identity ${^{k}{C_7}}$ combines with the next term ${^{k}{C_6}}$ from the original sum to form ${^{k+1}{C_7}}$. This process will continue until we reach the last term of the original sum, ${^{50}{C_6}}$. The combination immediately preceding the final step would be: ${^{49}{C_6}} + {^{49}{C_7}} = {^{50}{C_7}}$ And finally, combining ${^{50}{C_7}}$ with the last term from the original sum, ${^{50}{C_6}}$: ${^{50}{C_6}} + {^{50}{C_7}} = {^{51}{C_7}}$

4. Final Result (Method 1): After all the iterative combinations, the sum simplifies to ${^{51}{C_7}}$, and we still have the subtracted term $- {^{30}{C_7}}$. Therefore: $S = {}^{51}{C_7} - {}^{30}{C_7}$

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Method 2: Direct Application of the Hockey-stick Identity

The Hockey-stick Identity provides a more direct way to evaluate such sums. The form we use is: $\sum_{i=r}^{n} {^{i}{C_k}} = {^{n+1}{C_{k+1}}} - {^{r}{C_{k+1}}}$

1. Identify Parameters: From our expanded sum $S = {}^{30}{C_6} + {}^{31}{C_6} + \dots + {}^{50}{C_6}$, we can identify the parameters:

• The starting upper index, $r = 30$.
• The ending upper index, $n = 50$.
• The constant lower index, $k = 6$.

2. Apply the Identity: Substitute these values directly into the Hockey-stick Identity: $S = {^{50+1}{C_{6+1}}} - {^{30}{C_{6+1}}}$ $S = {^{51}{C_7}} - {^{30}{C_7}}$

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Tips for Success

• Recognize the pattern: Whenever you see a sum of combinations where the lower index is constant and the upper index is increasing (or decreasing), think of Pascal's Identity or the Hockey-stick Identity.
• Reorder the sum: If the upper index is decreasing, reordering it to an increasing sequence often makes applying the identities more straightforward.
• Catalyst Term: For iterative Pascal's Identity, remember to add and subtract a term (often the ${^n{C_{k+1}}}$ term for the smallest $n$ in the sum) to kickstart the chain.
• Beware of limits: Ensure you correctly identify the starting and ending indices ($r$ and $n$) when applying the Hockey-stick Identity, especially when the sum doesn't start from $k$.

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Summary and Key Takeaway

Both methods, iterative application of Pascal's Identity and direct use of the Hockey-stick Identity, lead to the same result. The Hockey-stick Identity is often quicker if remembered correctly. The value of the given summation is ${}^{51}{C_7} - {}^{30}{C_7}$.