Problem 73 - A Complete Resource Book in Mathematics for JEE Main 2017

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

Option (A) \(2^{2n}\).

Step 1: Understanding the Problem

We have a total of \(3n+1\) objects, consisting of \(n\) identical objects and \(2n+1\) distinct objects. We need to determine the number of ways to choose \(n\) objects from this collection.

Step 2: Count Choices for Identical Objects

Since all \(n\) identical objects must be considered the same, we can choose any number \(k\) (ranging from 0 to \(n\)) of these identical objects.

Step 3: Choose the Remaining Objects as Distinct

For each choice of \(k\) identical objects, we must choose the remaining \(n-k\) objects out of \(2n+1\) distinct objects. This is computed using combinations: \(\binom{2n+1}{n-k}\).

Step 4: Sum Over All Possible Choices for k

The total number of ways to choose \(n\) objects is the sum of all combinations for different values of \(k\): \[ \sum_{k=0}^{n} \binom{2n+1}{n-k} \]

Step 5: Apply the Combinatorial Identity

Using the combinatorial identity, the sum of combinations is equal to the sum of all subsets of the set of \(2n+1\) distinct objects: \[ \sum_{k=0}^{n} \binom{2n+1}{n-k} = 2^{2n} \]

Step 6: Determine the Correct Option

Since the total number of ways is \(2^{2n}\), the correct answer is option (A) \(2^{2n}\).

Key Concepts

Combinations

When solving problems related to choosing a subset of items from a larger set, combinations become very handy. Combinations are used when the order of selection does not matter. For instance, picking a committee of 5 people from a group of 20, where the order of selection doesn't impact the final committee, makes use of combinations. The mathematical notation for combinations to find the number of ways to choose \(r\) items from \(n\) items is given by:

• \(\binom{n}{r} = \frac{n!}{r! (n-r)!}\)

In the exercise at hand, combinations help determine how to select from the distinct objects available after accounting for the chosen identical ones. The approach involves considering different values for \(k\), the number of identical objects selected, and then using combinations for the remaining \(n-k\) objects out of the \(2n+1\) distinct items.

Combinatorial Identity

Combinatorial identities provide shortcuts for calculating sums involving combinations, which can be very practical in simplifying counting problems. One common combinatorial identity used is the identity for the sum of binomial coefficients. It states that if you sum the combinations of choosing \(r\) items from \(n\) items across various values of \(r\), the result equals the total number of subsets:

• \(\sum_{k=0}^{n} \binom{n}{k} = 2^n\)

In this exercise, we used a modified version where the sum of combinations for different values reduces the problem to a simpler form. We essentially calculated the sum for several ways to select remaining objects after choosing some identical ones. By employing this identity, the calculation is reduced to \(2^{2n}\), reflecting the power set concepts where each element can be either included or excluded, hence doubling with each addition.

Counting Problems

Counting problems involve determining the number of ways an event can occur or the number of items in some set. They are foundational in combinatorics and provide insight through structured problem solving. For this exercise, the key is to recognize the mix of identical and distinct objects, which introduces different complications than usual.
By decomposing the selection of objects into easier steps—first considering the identical ones and then the distinct—we systematically handle overlapping and missing cases. This ensures that all possibilities are accounted for. The critical step is always ensuring you have considered every potential selection path, which often involves breaking the problem down into countable segments and summing over distinct cases, as done here with different values for \(k\). Understanding and practicing these methods improve analytical skills necessary for overcoming more complex combinatorial challenges.