Problem 73 - A Complete Resource Book in Mathematics for JEE Main

Question

The number of ways of choosing \(n\) objects out of \((3 n+1)\) objects of which \(n\) are identical and \((2 n+1)\) are distinct, is (A) \(2^{2 n}\) (B) \(2^{2 n+1}\) (C) \(2^{2 n}-1\) (D) None of these

Step-by-Step Solution

Answer

The number of ways is \( 2^{2n} \), option (A).

Step 1: Understand the Problem

We need to choose \( n \) objects from a total of \( 3n + 1 \) objects, where \( n \) objects are identical and \( 2n + 1 \) objects are distinct. We will calculate the number of ways to do this.

Step 2: Consider Choosing Identical Objects

Since \( n \) objects are identical, when we choose \( k \) of these identical objects, we do not have to consider the distinct nature. The choices for these identical objects are from 0 to \( n \).

Step 3: Consider Choosing Distinct Objects

For each choice of \( n - k \) identical objects, we need to choose the remaining \( k \) objects from the \( 2n + 1 \) distinct objects. The choices here are given by \( \binom{2n+1}{k} \).

Step 4: Combine Choices

We sum over all the possible choices for \( k \), thus the total number of ways is \( \sum_{k=0}^{n} \binom{2n+1}{k} \).

Step 5: Evaluate the Combination Sum

The expression \( \sum_{k=0}^{n} \binom{2n+1}{k} \) by the binomial theorem represents half the terms from the binomial expansion of \( (1 + 1)^{2n+1} \), which is half of \( 2^{2n+1} \). This is calculated as: \( 2^{2n} \).

Step 6: Conclusion

Therefore, the number of ways to choose \( n \) objects from the given set is \( 2^{2n} \), matching option (A).

Key Concepts

Binomial Theorem

The binomial theorem is a fundamental concept in combinatorics which expresses the expanded form of a binomial expression raised to a power. Specifically, it expands expressions of the form \((x + y)^n\). The theorem states:
\[(x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k\]
In our exercise, this expansion is applied to calculate the combination sum for selecting objects. Each element of the sum, \(\binom{n}{k}\), signifies the number of ways to choose \(k\) elements from \(n\) total elements.
This exercise uses the binomial theorem to simplify the sum of combinations into a more manageable form, illustrated as calculating half of the expansion terms of \((1+1)^{2n+1}\), resulting in \(2^{2n}\).
Understanding this concept allows students to tackle problems that involve choosing subsets from sets, especially where distinctions between object types exist.

Combinatorial Analysis

Combinatorial analysis is a branch of mathematics dealing with the study of counting, arrangement, and combination of objects. The goal is to find how many ways arrangements can be made under given conditions.
In this exercise, we are tasked with finding how many ways we can select \(n\) objects from a mix of identical and distinct items out of \(3n + 1\) total items.
This method involves examining separate combinations one can make when objects are identical, versus when they are distinct.

• For identical objects, choosing any number between 0 to \(n\) is straightforward as each is indistinguishable from the other.
• For distinct objects, the solution requires calculating all potential combinations from the set of distinct items, adjusting parts of the set based on selections made from the identical category.

Combinatorial analysis is vital for systemizing the way we approach these types of counting problems, clearing ambiguity around combination choices.

Identical and Distinct Objects

The concept of identical and distinct objects is critical in permutation and combination problems. Identical objects are indistinguishable from each other, meaning swapping them does not create new arrangements. Distinct objects are unique, each item being different from the others.
In this exercise, understanding the difference in treatment between identical and distinct objects is key. Here’s the approach:

• For identical objects, any selection leads to the same outcome, hence choosing \(k\) out of these \(n\) objects results in a single choice for each \(k\).
• For distinct objects, however, each choice of selecting a group is unique and calculable using combinations. Thus, we use combinations to determine the potential arrangements from these \((2n+1)\) items.

By separating the decision-making process between identical and distinct classes, students can handle more complex problems where object identity plays a crucial role in determining outcomes.