Problem 85
Question
\({ }^{\mathrm{m}} C_{\mathrm{r}}+{ }^{\mathrm{m}} C_{\mathrm{r}-1} \cdot{ }^{\mathrm{n}} C_{1}+{ }^{\mathrm{m}} C_{\mathrm{r}-2} \cdot{ }^{\mathrm{n}} C_{2}+\ldots+{ }^{\mathrm{m}} C_{1} \cdot{ }^{\mathrm{n}} C_{\mathrm{r}-1}+{ }^{\mathrm{n}} C_{\mathrm{r}}=\) (A) \({ }^{\mathrm{m}+\mathrm{n}} C_{r-1}\) (B) \({ }^{\mathrm{m}+\mathrm{n}} \mathrm{C}_{\mathrm{r}}\) (C) \({ }^{\mathrm{m}+\mathrm{n}} C_{\mathrm{r}+1}\) (D) none of these
Step-by-Step Solution
Verified Answer
Option B: \({ }^{m+n} C_{r}\) is the correct answer.
1Step 1: Expand Binomial Coefficients
The problem involves a summation of binomial coefficients. The expression provided is a summation of individual terms combining coefficients from sets of size \(m\) and \(n\). This looks very similar to a well-known identity related to binomial coefficients known as the Hockey Stick or Christmas Stocking identity.
2Step 2: Recognize the Hockey Stick Identity
The hockey stick identity gives us that: \[{ }^{m} C_{r} + { }^{m} C_{r-1} \cdot{ }^{n} C_{1} + { }^{m} C_{r-2} \cdot{ }^{n} C_{2} + \ldots + { }^{n} C_{r} = { }^{m+n} C_{r}\] This means, the sum provided in the question is a straightforward application of the hockey stick identity. If each term of the sum is derived correctly, the whole expression would simplify according to the identity.
3Step 3: Simplify and Choose the Correct Answer
From the hockey stick identity, we can directly infer that the sum of the given terms \[{ }^{m} C_{r} + { }^{m} C_{r-1} \cdot{ }^{n} C_{1} + { }^{m} C_{r-2} \cdot{ }^{n} C_{2} + \ldots + { }^{n} C_{r} = { }^{m+n} C_{r}\]is equivalent to choosing \(r\) items out of \(m+n\) items. This is matching the option B.
Key Concepts
Binomial CoefficientHockey Stick IdentitySummation
Binomial Coefficient
When dealing with combinations and selections, the binomial coefficient plays a crucial role. It is a mathematical term that allows us to calculate how many ways we can choose a subset of items from a larger set. The binomial coefficient is represented as \( { }^{n}C_{r} \), which is read as \( n \) choose \( r \). This tells us how many ways we can choose \( r \) elements from a set of \( n \) elements, without regard to the order of selection.
Mathematically, the formula for the binomial coefficient is:
Using binomial coefficients, you can simplify and solve complex combinatorial problems efficiently. You calculate various possibilities and strategically choose the precise combinations needed for a specific scenario.
Mathematically, the formula for the binomial coefficient is:
- \( { }^{n} C_{r} = \frac{n!}{r!(n-r)!} \) where \(!\) denotes factorial.
Using binomial coefficients, you can simplify and solve complex combinatorial problems efficiently. You calculate various possibilities and strategically choose the precise combinations needed for a specific scenario.
Hockey Stick Identity
The Hockey Stick Identity is a fascinating principle in combinatorics that combines several binomial coefficients in a unique way. It gets its name from the shape that forms when these coefficients are placed in a Pascal's Triangle and these specific combinations are highlighted: they resemble a hockey stick. This identity is exceptionally useful when handling summations of binomial coefficients.
Formally, the Hockey Stick Identity is expressed as:
This identity simplifies the solution of problems that seem initially daunting by reducing them to a recognizable and computing-friendly form. It's a creative shortcut frequently helpful in both mathematical exercises and real-world applications.
Formally, the Hockey Stick Identity is expressed as:
- \( { }^{m} C_{r} + { }^{m} C_{r-1} \cdot { }^{n} C_{1} + { }^{m} C_{r-2} \cdot { }^{n} C_{2} + \ldots + { }^{m} C_{1} \cdot { }^{n} C_{r-1} + { }^{n} C_{r} = { }^{m+n} C_{r} \)
This identity simplifies the solution of problems that seem initially daunting by reducing them to a recognizable and computing-friendly form. It's a creative shortcut frequently helpful in both mathematical exercises and real-world applications.
Summation
Summation is a fundamental concept in mathematics that involves adding up a sequence of numbers or expressions. In the context of combinatorics, you often work with summations of binomial coefficients, such as in this exercise.
The goal in such scenarios is to find the total number of ways to achieve a particular result, by considering all possible combinations or paths that lead to it.
In this problem, the summation combines multiple terms that result from different combinations of choosing items from two sets. By applying the Hockey Stick Identity, we use summation to sum the products of the binomial coefficients in a manner that simplifies our calculations.
Summation in combinatorics often involves:
The goal in such scenarios is to find the total number of ways to achieve a particular result, by considering all possible combinations or paths that lead to it.
In this problem, the summation combines multiple terms that result from different combinations of choosing items from two sets. By applying the Hockey Stick Identity, we use summation to sum the products of the binomial coefficients in a manner that simplifies our calculations.
Summation in combinatorics often involves:
- Recognizing patterns or identities that simplify extensive calculations.
- Ensuring that terms align with known combinatorial identities.
Other exercises in this chapter
Problem 83
For all \(n \in N\), the integer just above \((\sqrt{3}+1)^{2 n}\) is divisible by (A) \(2^{\mathrm{n}+1}\) (B) \(2^{\mathrm{n}}+1\) (C) \(2^{\mathrm{n}+1}+1\)
View solution Problem 84
If \(C_{0}, C_{1}, C_{2}, \ldots, C_{\mathrm{n}}\) be the coefficients in the expansion of \((1+x)^{\mathrm{n}}\), then \(\frac{2^{2} \cdot C_{0}}{1 \cdot 2}+\f
View solution Problem 86
If \(a, b, c\) and \(d\) are any four consecutive coefficients of any binomial expansion, then \(\frac{a+b}{a}, \frac{b+c}{b}, \frac{c+d}{c}\) are (A) A.P. (B)
View solution Problem 87
The last two digits of the number \(3^{400}\) are (A) 38 (B) 27 (C) 01 (D) none of these
View solution