The value of \(^{47}C_4 + \displaystyle\sum_{r=1}^5 {^{52-r}C_3}\) is equal to:
Evaluating the Value of a Combinations Summation
The problem asks us to find the value of the expression: \(^{47}C_4 + \displaystyle\sum_{r=1}^5 {^{52-r}C_3}\).
First, let's expand the summation part of the expression. The sum runs from \(r=1\) to \(r=5\). When \(r=1\), the term is \(^{52-1}C_3 = {^{51}C_3}\). When \(r=2\), the term is \(^{52-2}C_3 = {^{50}C_3}\). When \(r=3\), the term is \(^{52-3}C_3 = {^{49}C_3}\). When \(r=4\), the term is \(^{52-4}C_3 = {^{48}C_3}\). When \(r=5\), the term is \(^{52-5}C_3 = {^{47}C_3}\).
So, the summation is:
\(\displaystyle\sum_{r=1}^5 {^{52-r}C_3} = {^{51}C_3} + {^{50}C_3} + {^{49}C_3} + {^{48}C_3} + {^{47}C_3}\)
Now, let's rewrite the original expression with the expanded summation:
\(^{47}C_4 + {^{47}C_3} + {^{48}C_3} + {^{49}C_3} + {^{50}C_3} + {^{51}C_3}\)
Applying Pascal's Identity for Combinations
To simplify this expression, we can use Pascal's Identity, which states:
\(^nC_r + ^nC_{r-1} = ^{n+1}C_r\)
Let's group the terms in our expression and apply this identity step-by-step:
Start with the first two terms: \(^{47}C_4 + {^{47}C_3}\). Using Pascal's Identity with \(n=47\) and \(r=4\), we have \(^{47}C_4 + {^{47}C_{4-1}} = ^{47+1}C_4\), which simplifies to \(^{48}C_4\).
Now the expression becomes:
\(^{48}C_4 + {^{48}C_3} + {^{49}C_3} + {^{50}C_3} + {^{51}C_3}\)
Next, group the first two terms again: \(^{48}C_4 + {^{48}C_3}\). Using Pascal's Identity with \(n=48\) and \(r=4\), we have \(^{48}C_4 + {^{48}C_{4-1}} = ^{48+1}C_4\), which simplifies to \(^{49}C_4\).
The expression is now:
\(^{49}C_4 + {^{49}C_3} + {^{50}C_3} + {^{51}C_3}\)
Repeat the process: \(^{49}C_4 + {^{49}C_3}\). Using Pascal's Identity with \(n=49\) and \(r=4\), we get \(^{49+1}C_4 = {^{50}C_4}\).
The expression is now:
\(^{50}C_4 + {^{50}C_3} + {^{51}C_3}\)
Repeat again: \(^{50}C_4 + {^{50}C_3}\). Using Pascal's Identity with \(n=50\) and \(r=4\), we get \(^{50+1}C_4 = {^{51}C_4}\).
The expression is now:
\(^{51}C_4 + {^{51}C_3}\)
Finally, apply Pascal's Identity one last time: \(^{51}C_4 + {^{51}C_3}\). Using Pascal's Identity with \(n=51\) and \(r=4\), we get \(^{51+1}C_4 = {^{52}C_4}\).
So, the value of the given expression is \(^{52}C_4\).
Comparing with Options
Let's compare our result with the given options:
- \(^{52}C_4\)
- \(^{51}C_4\)
- \(^{51}C_3\)
- None of these
Our calculated value \(^{52}C_4\) matches option 1.
Step-by-Step Simplification
| Step | Expression | Identity Applied | Result |
|---|---|---|---|
| 1 | \(^{47}C_4 + {^{47}C_3} + {^{48}C_3} + {^{49}C_3} + {^{50}C_3} + {^{51}C_3}\) | \(^{47}C_4 + {^{47}C_3} = ^{48}C_4\) | \(^{48}C_4 + {^{48}C_3} + {^{49}C_3} + {^{50}C_3} + {^{51}C_3}\) |
| 2 | \(^{48}C_4 + {^{48}C_3} + {^{49}C_3} + {^{50}C_3} + {^{51}C_3}\) | \(^{48}C_4 + {^{48}C_3} = ^{49}C_4\) | \(^{49}C_4 + {^{49}C_3} + {^{50}C_3} + {^{51}C_3}\) |
| 3 | \(^{49}C_4 + {^{49}C_3} + {^{50}C_3} + {^{51}C_3}\) | \(^{49}C_4 + {^{49}C_3} = ^{50}C_4\) | \(^{50}C_4 + {^{50}C_3} + {^{51}C_3}\) |
| 4 | \(^{50}C_4 + {^{50}C_3} + {^{51}C_3}\) | \(^{50}C_4 + {^{50}C_3} = ^{51}C_4\) | \(^{51}C_4 + {^{51}C_3}\) |
| 5 | \(^{51}C_4 + {^{51}C_3}\) | \(^{51}C_4 + {^{51}C_3} = ^{52}C_4\) | \(^{52}C_4\) |
Revision Table: Key Concepts in Combinations
| Concept | Description | Formula |
|---|---|---|
| Combination | The number of ways to choose \(r\) items from a set of \(n\) distinct items without regard to the order of selection. | \(^nC_r = \frac{n!}{r!(n-r)!}\) |
| Pascal's Identity | A fundamental identity relating binomial coefficients. It forms the basis of Pascal's Triangle. | \(^nC_r + ^nC_{r-1} = ^{n+1}C_r\) |
| Summation Notation | A compact way to represent the sum of a sequence of terms. | \(\displaystyle\sum_{i=k}^m a_i = a_k + a_{k+1} + \dots + a_m\) |
Additional Information: Properties of Combinations and Pascal's Triangle
Combinations, denoted by \(^nC_r\) or \(\binom{n}{r}\), are also known as binomial coefficients. They represent the coefficients in the binomial expansion of \((x+y)^n\). For example, \((x+y)^4 = ^4C_0 x^4 y^0 + ^4C_1 x^3 y^1 + ^4C_2 x^2 y^2 + ^4C_3 x^1 y^3 + ^4C_4 x^0 y^4\).
Pascal's Identity (\(^nC_r + ^nC_{r-1} = ^{n+1}C_r\)) has a simple combinatorial interpretation: To choose \(r\) elements from a set of \(n+1\) elements (\(^{n+1}C_r\)), you can either choose the first element and \(r-1\) elements from the remaining \(n\) (\(^nC_{r-1}\)), or not choose the first element and choose \(r\) elements from the remaining \(n\) (\(^nC_r\)).
Pascal's Triangle is a triangular arrangement of binomial coefficients. Each number in the triangle is the sum of the two numbers directly above it. This directly illustrates Pascal's Identity. The \(n\)-th row of Pascal's Triangle contains the values \(^nC_0, ^nC_1, \dots, ^nC_n\).
The identity used in this problem, \(\displaystyle\sum_{k=m}^n {^kC_r} = {^{n+1}C_{r+1}} - {^mC_{r+1}}\) for \(r \ge 0, n \ge m \ge r\), is related but the step-by-step application of Pascal's identity is more intuitive for this specific sum structure. The given sum is \(^{47}C_3 + {^{48}C_3} + {^{49}C_3} + {^{50}C_3} + {^{51}C_3}\). Let's add \(^{47}C_4\) and rewrite the sum slightly: \(^{47}C_4 + {^{47}C_3} + {^{48}C_3} + {^{49}C_3} + {^{50}C_3} + {^{51}C_3}\). This structure is perfectly suited for the repeated application of \(^nC_r + ^nC_{r-1} = ^{n+1}C_r\).
Understanding these basic identities and properties of combinations is crucial for solving more complex problems in combinatorics and probability.
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