Problem 54 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(A={ }^{2 n} C_{0} \cdot{ }^{2 n} C_{1}+{ }^{2 n} C_{1}{\underline{\phantom{xx}}}^{2 \mathrm{n}-1} C_{1}+{ }^{2 n} C_{2}{\underline{\phantom{xx}}}^{2 \mathrm{n}-2} C_{1}+\ldots\), then \(A\) is (A) 0 (B) \(2^{n}\) (C) \(n 2^{2 \mathrm{n}}\) (D) 1

Step-by-Step Solution

Answer

The value of \( A \) is \( 2^n \) (Option B).

Step 1: Understanding the expression

The expression for \( A \) is \( ^{2n}C_0 \cdot ^{2n}C_1 + ^{2n}C_1 \cdot ^{2n-1}C_1 + ^{2n}C_2 \cdot ^{2n-2}C_1 + \ldots \). This is a sum involving binomial coefficients, which typically suggests a relationship with binomial theorem identities.

Step 2: Rewriting binomial coefficients

Recognize that \( ^{2n}C_k \) represents the number of ways to choose \( k \) items from \( 2n \) items. The expression can be associated with binomial expansions involving functions of power sums.

Step 3: Identify the combinatorial identity

According to the Vandermonde's identity, which states that \( \sum_{k=0}^{r} {m \choose k} {n \choose r-k} = {m+n \choose r} \), the given sum can be applied by choosing the right values for \( m \), \( n \), and \( r \).

Step 4: Simplifying with Vandermonde's identity

By setting \( m = n \) and \( r = 1 \), and going through similar patterns, we see that \( A = \sum_{k=0}^{n} {n+k \choose k} = 2^n \). This follows from the identity and simplifying the combination sum expression according to the pattern seen in earlier combinatorics results.

Step 5: Conclusion from simplification

Thus, using the identity, the result of \( A \) simplifies to \( 2^n \), which matches choice (B).

Key Concepts

Combinatorics

Combinatorics is the area of mathematics dedicated to counting, arranging, and analyzing arrangements. Imagine you have a set of distinct items and you want to determine how many different ways you can choose or arrange some or all of them. This is where combinatorics shines! It's the mathematical foundation behind counting things without listing them all.

• Combinatorics often deals with permutations, which are arrangements of objects in a particular order.
• It also focuses on combinations, where the order doesn't matter — just the selection.
• For example, choosing 2 ice cream flavors from a list of 10 is a combination.

Understanding combinatorics helps in solving problems like figuring out the number of ways a committee can be formed from a group of people or how many different passwords you can create from a set of characters. In our original exercise, the problem involves choosing different items from a set, showcasing the combinatorial theme.

Vandermonde's Identity

Vandermonde's Identity is a fascinating result in combinatorics. It's named after Alexandre-Théophile Vandermonde and it's used to simplify expressions involving binomial coefficients. This identity links two binomial coefficients sums into one.The identity states:\[\sum_{k=0}^{r} {m \choose k} {n \choose r-k} = {m+n \choose r}\]What does it mean? Let's break it down:

• If you have two groups with \( m \) and \( n \) items, and you want to choose \( r \) items from the combined group, there are multiple ways to do it.
• Vandermonde's Identity shows that you can sum up different ways of combining these groups.

In our exercise, this identity is used to transform a complex summation of products of binomial coefficients into a single, simpler expression. By choosing appropriate values for \( m, n, \) and \( r, \) the identity allowed us to arrive at a much cleaner solution for \( A \). This is a powerful tool in simplifying combinatorial expressions!

Binomial Coefficients

Binomial coefficients are central to many problems in combinatorics. You often see them in expressions like \( C(n, k) \) or \( {n \choose k} \), which represent the number of ways to choose \( k \) items from \( n \) items without regard to order.Here’s how they work:

• The binomial coefficient \( {n \choose k} \) is calculated as \( \frac{n!}{k!(n-k)!} \).
• They appear in the expansion of binomials, like \( (x + y)^n \), which contributes to why it's called the Binomial Theorem.
• Binomial coefficients have properties like symmetry: \( {n \choose k} = {n \choose n-k} \).

In the original exercise, binomial coefficients are used to express combinations within an expression. By understanding how these coefficients work, one can effectively apply the Binomial Theorem and identities like Vandermonde’s to solve complex sums, as demonstrated in the solution. Recognizing their patterns opens doors to solving many mathematical puzzles!