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Animated Solution for Mathematics - Binomial Theorem: The value of $^{50}C_4 + \sum_{r=1}^6 ^{56-r}C_3$ is

Q: How do I know when to use Pascal's Identity versus other identities like the Hockey Stick Identity?

Great question! Pascal's Identity ($^{n}C_{r} + ^{n}C_{r-1} = ^{n+1}C_{r}$) is your go-to when you have two terms with the same upper index and consecutive lower indices. The Hockey Stick Identity is usually applied when you have a sum of terms where the lower index is constant and the upper index increases. Here, we are essentially performing a 'cascading' Pascal's Identity, which leads us to the same result as the Hockey Stick Identity.

Select Answer:

Visualized Solution

Analyzing the Given Series

Given expression: $^{50} C_{4} + \sum_{r = 1}^{6}^{56 - r} C_{3}$
We have a single binomial coefficient added to a summation of six terms.
To understand the pattern, we must first expand the summation.

Expanding the Summation Terms

Let's substitute $r = 1, 2, 3, 4, 5, 6$ into $^{56 - r} C_{3}$ .
For $r = 1$ : $^{55} C_{3}$
For $r = 2$ : $^{54} C_{3}$
... down to $r = 6$ : $^{50} C_{3}$
The expanded series is: $^{50} C_{4} +^{55} C_{3} +^{54} C_{3} +^{53} C_{3} +^{52} C_{3} +^{51} C_{3} +^{50} C_{3}$

Rearranging for Pattern Recognition

To apply standard identities, we group terms with the same upper index.
Let's write the series in reverse order of the summation terms.
Rearranged: $(^{50} C_{4} +^{50} C_{3}) +^{51} C_{3} +^{52} C_{3} +^{53} C_{3} +^{54} C_{3} +^{55} C_{3}$

The Logic Bridge: Pascal's Identity

Identify the core tool: Pascal's Identity
Formula: $^{n} C_{r} +^{n} C_{r - 1} =^{n + 1} C_{r}$
Condition 1: Upper indices must be identical ( $n$ ).
Condition 2: Lower indices must be consecutive ( $r$ and $r - 1$ ).

Atomic Compute: First Addition

Apply identity to the first group: $^{50} C_{4} +^{50} C_{3}$
Here, $n = 50$ , and the lower indices $4$ and $3$ are consecutive.
Result: $^{51} C_{4}$
The series becomes: $^{51} C_{4} +^{51} C_{3} +^{52} C_{3} + \dots +^{55} C_{3}$

Atomic Compute: Second Addition

Now, group the new term with the next term in the series: $^{51} C_{4} +^{51} C_{3}$
Here, $n = 51$ , and lower indices are $4$ and $3$ .
Result: $^{52} C_{4}$
The series becomes: $^{52} C_{4} +^{52} C_{3} +^{53} C_{3} + \dots +^{55} C_{3}$

Atomic Compute: Third Addition

Continue the chain reaction: $^{52} C_{4} +^{52} C_{3}$
Result: $^{53} C_{4}$
The series becomes: $^{53} C_{4} +^{53} C_{3} +^{54} C_{3} +^{55} C_{3}$

Atomic Compute: Fourth Addition

Next pair: $^{53} C_{4} +^{53} C_{3}$
Result: $^{54} C_{4}$
The series becomes: $^{54} C_{4} +^{54} C_{3} +^{55} C_{3}$

Atomic Compute: Fifth Addition

Next pair: $^{54} C_{4} +^{54} C_{3}$
Result: $^{55} C_{4}$
The series becomes: $^{55} C_{4} +^{55} C_{3}$

Final Addition Step

The final pair: $^{55} C_{4} +^{55} C_{3}$
Here, $n = 55$ , and lower indices are $4$ and $3$ .
Result: $^{56} C_{4}$
The entire summation has collapsed into a single term.

Final Result and Takeaway

Final Answer: $^{56} C_{4}$
Key Takeaway: Pascal's identity ( $^{n} C_{r} +^{n} C_{r - 1} =^{n + 1} C_{r}$ ) creates a cascading effect when applied to a series of binomial coefficients with sequential upper indices.
Pro Tip: Always look for a starting pair with identical upper indices to trigger the chain reaction.

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The Sigma Insight: Properties of Binomial Coefficients

Solution Diagram

Analyzing the Setup

The problem asks us to evaluate the expression:

^{50}C_4 + \sum_{r=1}^{6} ^{56-r}C_3

At first glance, it is easy to feel overwhelmed. You might be tempted to grind through the factorials, but in the world of JEE Advanced, the path to victory is pattern recognition.

Unveiling the Hidden Sequence

We must peel back the layers of the summation. The expression

^{56 - r} C_{3}

represents a sequence of terms.

When we expand the summation for

r = 1

r = 6

, we obtain:

^{55} C_{3} +^{54} C_{3} +^{53} C_{3} +^{52} C_{3} +^{51} C_{3} +^{50} C_{3}

The full expression now appears as:

^{50} C_{4} + (^{55} C_{3} +^{54} C_{3} +^{53} C_{3} +^{52} C_{3} +^{51} C_{3} +^{50} C_{3})

The Power of Pascal's Identity

The core of our strategy is Pascal's Identity, which states:

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

This identity acts as a bridge, allowing us to combine two terms into a single, unified result. We begin by grouping the last term of the summation with the initial term:

^{50} C_{4} +^{50} C_{3} =^{51} C_{4}

The Domino Effect

Now, we observe a cascading effect. We take our new term

^{51} C_{4}

and combine it with the next term in the sequence,

^{51} C_{3}

^{51} C_{4} +^{51} C_{3} =^{52} C_{4}

Continuing this process, the dominoes fall in sequence:

^{52} C_{4} +^{52} C_{3} =^{53} C_{4}

^{53} C_{4} +^{53} C_{3} =^{54} C_{4}

^{54} C_{4} +^{54} C_{3} =^{55} C_{4}

^{55} C_{4} +^{55} C_{3} =^{56} C_{4}

Final Calculation

By applying the identity repeatedly, the entire expression simplifies into one elegant term. The final result is:

^{56} C_{4}

Calculating the numerical value:

\frac{56 \times 55 \times 54 \times 53}{4 \times 3 \times 2 \times 1} = 369, 600

The essence of JEE mathematics is seeing this hidden structure. When you approach these problems, look for the identity that wants to be used and trust in the process.

Common Traps & FAQs

Why can't I just calculate the values of

^{50} C_{4}

and the summation terms? It seems like a lot of work, but is it impossible?

While technically possible, it is a trap! In JEE Advanced, time is your most precious currency. Calculating

^{50} C_{4}

involves large factorials. The problem is designed to test your ability to recognize the 'Pascal's Identity' pattern, not your arithmetic speed. Always look for the algebraic shortcut first.

How do I know when to use Pascal's Identity versus other identities like the Hockey Stick Identity?

Great question! Pascal's Identity (

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

) is your go-to when you have two terms with the same upper index and consecutive lower indices. The Hockey Stick Identity is usually applied when you have a sum of terms where the lower index is constant and the upper index increases. Here, we are essentially performing a 'cascading' Pascal's Identity, which leads us to the same result as the Hockey Stick Identity.

I got stuck because I didn't know where to start the grouping. Does the order matter?

The order is everything! By expanding the summation, you revealed that the term

^{50} C_{3}

was hidden at the end. By placing it next to

^{50} C_{4}

, you created the perfect condition for Pascal's Identity. If you don't see the pattern, always write out the first and last few terms of the summation to visualize the sequence.

Animated Solution for Mathematics - Binomial Theorem: The value of $^{50}C_4 + \sum_{r=1}^6 ^{56-r}C_3$ is

Select Answer:

Visualized Solution

Analyzing the Given Series

Expanding the Summation Terms

Rearranging for Pattern Recognition

The Logic Bridge: Pascal's Identity

Atomic Compute: First Addition

Atomic Compute: Second Addition

Atomic Compute: Third Addition

Atomic Compute: Fourth Addition

Atomic Compute: Fifth Addition

Final Addition Step

Final Result and Takeaway

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

Unveiling the Hidden Sequence

The Power of Pascal's Identity

The Domino Effect

Final Calculation

Similar Questions

The value of $\sum_{r=0}^{20} ^{50-r}C_6$ is equal to:

$\sum_{k = 0}^{6}^{51 - k} C_{3}$ is equal to

The value $\sum_{r = 0}^{22}^{22} C_{r} \cdot^{23} C_{r}$ is

The value of $\frac{^{100} C _{50}}{51} + \frac{^{100} C _{51}}{52} + \dots + \frac{^{100} C _{100}}{101}$ is :

The value of the expression $^{47} C_{4} + \sum_{j = 1}^{5}^{52 - j} C_{3}$ is equal to

The value of $\frac{1}{1 ! 50 !} + \frac{1}{3 ! 48 !} + \frac{1}{5 ! 46 !} + \dots + \frac{1}{49 ! 2 !} + \frac{1}{51 ! 1 !}$ is

The value of $(^{21} C_{1} -^{10} C_{1}) + (^{21} C_{2} -^{10} C_{2}) + (^{21} C_{3} -^{10} C_{3}) + (^{21} C_{4} -^{10} C_{4}) + \dots + (^{21} C_{10} -^{10} C_{10})$ is:

The value of $\sum_{r = 0}^{6} (^{6} C_{r} \cdot^{6} C_{6 - r})$ is equal to :

If $C_{r} =^{25} C_{r}$ and $C_{0} + 5 \cdot C_{1} + 9 \cdot C_{2} + \dots + 101 \cdot C_{25} = 2^{25} \cdot k$ then $k$ is equal to__________.

The sum of the series $2 .^{20} C_{0} + 5 .^{20} C_{1} + 8 .^{20} C_{2} + 11 .^{20} C_{3} + ... + 62 .^{20} C_{20}$ is equal to :

Animated Solution for Mathematics - Binomial Theorem: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of ^{50}C_4 + \sum_{r=1}^6 ^{56-r}C_3 is

Select Answer:

Visualized Solution

Analyzing the Given Series

Expanding the Summation Terms

Rearranging for Pattern Recognition

The Logic Bridge: Pascal's Identity

Atomic Compute: First Addition

Atomic Compute: Second Addition

Atomic Compute: Third Addition

Atomic Compute: Fourth Addition

Atomic Compute: Fifth Addition

Final Addition Step

Final Result and Takeaway

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

Unveiling the Hidden Sequence

The Power of Pascal's Identity

The Domino Effect

Final Calculation

Similar Questions

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Animated Solution for Mathematics - Binomial Theorem: The value of $^{50}C_4 + \sum_{r=1}^6 ^{56-r}C_3$ is

The value of $\sum_{r=0}^{20} ^{50-r}C_6$ is equal to:

$\sum_{k = 0}^{6}^{51 - k} C_{3}$ is equal to

The value $\sum_{r = 0}^{22}^{22} C_{r} \cdot^{23} C_{r}$ is

The value of $\frac{^{100} C _{50}}{51} + \frac{^{100} C _{51}}{52} + \dots + \frac{^{100} C _{100}}{101}$ is :

The value of the expression $^{47} C_{4} + \sum_{j = 1}^{5}^{52 - j} C_{3}$ is equal to

The value of $\frac{1}{1 ! 50 !} + \frac{1}{3 ! 48 !} + \frac{1}{5 ! 46 !} + \dots + \frac{1}{49 ! 2 !} + \frac{1}{51 ! 1 !}$ is

The value of $(^{21} C_{1} -^{10} C_{1}) + (^{21} C_{2} -^{10} C_{2}) + (^{21} C_{3} -^{10} C_{3}) + (^{21} C_{4} -^{10} C_{4}) + \dots + (^{21} C_{10} -^{10} C_{10})$ is:

The value of $\sum_{r = 0}^{6} (^{6} C_{r} \cdot^{6} C_{6 - r})$ is equal to :

If $C_{r} =^{25} C_{r}$ and $C_{0} + 5 \cdot C_{1} + 9 \cdot C_{2} + \dots + 101 \cdot C_{25} = 2^{25} \cdot k$ then $k$ is equal to__________.

The sum of the series $2 .^{20} C_{0} + 5 .^{20} C_{1} + 8 .^{20} C_{2} + 11 .^{20} C_{3} + ... + 62 .^{20} C_{20}$ is equal to :