Key Concept: Pascal's Identity (Combinatorial Identity)

The fundamental principle for solving this problem is Pascal's Identity, a cornerstone in combinatorics. It states that for any non-negative integers $n$ and $r$ where $n \ge r \ge 1$:

${^n{C_r}} + {^n{C_{r - 1}}} = {^{n + 1}{C_r}}$

Why is this identity important? Pascal's Identity has a clear and intuitive combinatorial meaning. Consider forming a committee of $r$ members from a group of $n+1$ distinct individuals. Let's pick one specific individual, say 'X', from the $n+1$ people.

• Case 1: Individual 'X' is included in the committee. If 'X' is chosen, we still need to select $r-1$ more members from the remaining $n$ individuals. The number of ways to do this is ${^n{C_{r-1}}}$.
• Case 2: Individual 'X' is not included in the committee. If 'X' is excluded, we must choose all $r$ members from the remaining $n$ individuals. The number of ways to do this is ${^n{C_r}}$. Since these two cases are mutually exclusive and exhaustive (they cover all possibilities), their sum must equal the total number of ways to choose $r$ members from $n+1$ individuals, which is ${^{n+1}{C_r}}$. This identity is also famously depicted in Pascal's Triangle, where each number is the sum of the two numbers directly above it.

Understanding the Problem Statement

We are asked to evaluate the following expression: ${}^{50}{C_4} + \sum\limits_{r = 1}^6 {^{56 - r}} {C_3}$

Step 1: Expand the Summation To clearly identify all the terms and prepare them for the application of Pascal's Identity, our first step is to expand the summation part, $\sum\limits_{r = 1}^6 {^{56 - r}} {C_3}$.

Let's list the terms by substituting each value of $r$ from 1 to 6:

• For $r=1$: ${}^{56 - 1}{C_3} = {}^{55}{C_3}$
• For $r=2$: ${}^{56 - 2}{C_3} = {}^{54}{C_3}$
• For $r=3$: ${}^{56 - 3}{C_3} = {}^{53}{C_3}$
• For $r=4$: ${}^{56 - 4}{C_3} = {}^{52}{C_3}$
• For $r=5$: ${}^{56 - 5}{C_3} = {}^{51}{C_3}$
• For $r=6$: ${}^{56 - 6}{C_3} = {}^{50}{C_3}$

Step 2: Rewrite and Rearrange the Expression Now, we substitute these expanded terms back into the original expression. To facilitate the application of Pascal's Identity, it's beneficial to arrange the terms in a specific order. We'll place ${}^{50}{C_4}$ adjacent to ${}^{50}{C_3}$ to initiate the chain of simplifications.

The expression initially becomes: ${}^{50}{C_4} + {}^{55}{C_3} + {}^{54}{C_3} + {}^{53}{C_3} + {}^{52}{C_3} + {}^{51}{C_3} + {}^{50}{C_3}$ Rearranging the terms in ascending order of the upper index 'n' for the $C_3$ terms, and strategically placing ${}^{50}{C_4}$: ${}^{50}{C_4} + {}^{50}{C_3} + {}^{51}{C_3} + {}^{52}{C_3} + {}^{53}{C_3} + {}^{54}{C_3} + {}^{55}{C_3}$

Step-by-Step Iterative Application of Pascal's Identity (The "Hockey-stick" Pattern)

The core strategy here is to repeatedly apply Pascal's Identity to progressively simplify the sum. This creates a "chain reaction" where each application produces a new term that then combines with the next term in the sequence. This specific pattern of summation is famously known as the Hockey-stick Identity.

1. First Application: We begin by combining the first two terms: ${}^{50}{C_4} + {}^{50}{C_3}$.

• Why this pair? This pair perfectly matches the structure of Pascal's Identity ${^n{C_r}} + {^n{C_{r - 1}}} = {^{n + 1}{C_r}}$, where $n=50$, $r=4$, and $r-1=3$. Applying the identity: $\left( {}^{50}{C_4} + {}^{50}{C_3} \right) = {}^{50+1}{C_4} = {}^{51}{C_4}$ The expression now simplifies to: ${}^{51}{C_4} + {}^{51}{C_3} + {}^{52}{C_3} + {}^{53}{C_3} + {}^{54}{C_3} + {}^{55}{C_3}$ Notice how the newly formed term, ${}^{51}{C_4}$, is now perfectly positioned to combine with the next $C_3$ term, ${}^{51}{C_3}$.

2. Second Application: Next, we combine ${}^{51}{C_4}$ with ${}^{51}{C_3}$.

• Why this pair? This continues the established pattern, fitting Pascal's Identity with $n=51$, $r=4$, and $r-1=3$. Applying the identity: $\left( {}^{51}{C_4} + {}^{51}{C_3} \right) = {^{51+1}}{C_4} = {}^{52}{C_4}$ The expression becomes: ${}^{52}{C_4} + {}^{52}{C_3} + {}^{53}{C_3} + {}^{54}{C_3} + {}^{55}{C_3}$

3. Third Application: We continue the pattern by combining ${}^{52}{C_4}$ with ${}^{52}{C_3}$.

• Why this pair? Following the same logic, $n=52$, $r=4$, and $r-1=3$. Applying the identity: $\left( {}^{52}{C_4} + {}^{52}{C_3} \right) = {^{52+1}}{C_4} = {}^{53}{C_4}$ The expression is now: ${}^{53}{C_4} + {}^{53}{C_3} + {}^{54}{C_3} + {}^{55}{C_3}$

4. Fourth Application: Combine ${}^{53}{C_4}$ with ${}^{53}{C_3}$.

• Why this pair? Continuing the established pattern: $n=53$, $r=4$, $r-1=3$. Applying the identity: $\left( {}^{53}{C_4} + {}^{53}{C_3} \right) = {^{53+1}}{C_4} = {}^{54}{C_4}$ The expression becomes: ${}^{54}{C_4} + {}^{54}{C_3} + {}^{55}{C_3}$

5. Fifth Application: Combine ${}^{54}{C_4}$ with ${}^{54}{C_3}$.

• Why this pair? Adhering to the pattern: $n=54$, $r=4$, $r-1=3$. Applying the identity: $\left( {}^{54}{C_4} + {}^{54}{C_3} \right) = {^{54+1}}{C_4} = {}^{55}{C_4}$ The expression simplifies to: ${}^{55}{C_4} + {}^{55}{C_3}$

6. Final Application: Finally, we combine ${}^{55}{C_4}$ with ${}^{55}{C_3}$.

• Why this pair? This is the last pair that fits Pascal's Identity: $n=55$, $r=4$, $r-1=3$. Applying the identity: $\left( {}^{55}{C_4} + {}^{55}{C_3} \right) = {^{55+1}}{C_4} = {}^{56}{C_4}$

Final Result

The value of the given expression is ${}^{56}{C_4}$.

Important Tips and Common Pitfalls to Avoid

1. The Hockey-stick Identity: This problem is a direct application of the "Hockey-stick Identity" (also known as the Christmas stocking identity). The general form is: $\sum_{i=r}^n {^iC_r} = {^{n+1}C_{r+1}}$ In our case, the sum can be written as ${}^{50}{C_4} + \sum_{k=50}^{55} {^kC_3}$. Using the Hockey-stick identity for the sum part: $\sum_{k=50}^{55} {^kC_3} = \left( \sum_{k=3}^{55} {^kC_3} \right) - \left( \sum_{k=3}^{49} {^kC_3} \right)$ $= {^{55+1}C_{3+1}} - {^{49+1}C_{3+1}} = {^{56}C_4} - {^{50}C_4}$ Substituting this back into the original expression: ${}^{50}{C_4} + \left( {^{56}C_4} - {^{50}C_4} \right) = {}^{56}{C_4}$ This confirms the result obtained through iterative application of Pascal's Identity. The iterative method is essentially how the Hockey-stick identity is derived.

2. Correctly Applying Pascal's Identity: Always ensure that the two terms you are combining perfectly match the form ${^n{C_r}} + {^n{C_{r - 1}}}$. This means the upper index ($n$) must be identical for both terms, and the lower indices ($r$ and $r-1$) must differ by exactly one. A common mistake is to misidentify $n$ or $r$, or to try to combine terms that don't fit this precise structure.

3. Strategic Rearrangement: While addition is commutative, arranging the terms in a specific order (e.g., placing $C_r$ next to $C_{r-1}$ with the same $n$) makes the pattern for applying Pascal's Identity immediately obvious and simplifies the calculation process.

Summary and Key Takeaway

This problem is a classic example demonstrating the power and elegance of Pascal's Identity in simplifying sums of binomial coefficients. By systematically expanding the summation and then iteratively applying Pascal's Identity, a seemingly complex sum can be reduced to a single binomial coefficient. This technique is a fundamental skill in combinatorics and is frequently tested in competitive examinations like JEE. Understanding the underlying Hockey-stick Identity provides a faster way to arrive at the solution once the pattern is recognized.