Problem 46 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The number of ways of choosing 10 objects out of 31 objects of which 10 are identical and the remaining 21 are distinct is: (a) \(2^{20}-1\) (b) \(2^{23}\) (c) \(2^{20}\) (d) \(2^{x}+1\)

Step-by-Step Solution

Answer

The answer is (c) \(2^{20}\).

Step 1: Understand the Problem

We need to choose 10 objects out of 31, where 10 are identical and 21 are distinct. The task is to find the number of ways to do this.

Step 2: Analyze the Distinct Part

Since 10 objects are identical, let's focus first on selecting from the 21 distinct objects. We need to consider combinations of 0 to 10 distinct objects since the sum of identical and distinct objects must be 10.

Step 3: Calculate Combinations

For each possible number from 0 to 10 distinct objects chosen, calculate the ways using combinations. E.g., choosing r distinct objects uses the formula \( \binom{21}{r} \).

Step 4: Incorporate Identical Objects

For each combination, see how many identical objects are needed to make a total of 10 objects. E.g., if 3 distinct objects are chosen, then choose 7 identical objects.

Step 5: Summing Combinations

Compute the total number of combinations by summing \( \binom{21}{r} \) over all valid values of r (from 0 to 10). This is equivalent to computing \( \sum_{r=0}^{10} \binom{21}{r} \).

Step 6: Apply Binomial Theorem

The sum from step 5 follows the expansion of \((1+1)^{21}\) which gives \(2^{21}\), but we only sum up to r=10 so subtract the cases higher than 10.

Step 7: Subtraction Calculation

The sum of selecting more than 10 objects is \( \sum_{r=11}^{21} \binom{21}{r} \), which due to symmetry is equivalent to \( \sum_{r=0}^{10} \binom{21}{r} \). Therefore, \(2^{21}\) covers all combinations, we effectively sum \(2^{21-1} = 2^{20}\).

Key Concepts

Binomial Theorem

The binomial theorem is a powerful tool in combinatorics that allows us to expand expressions that are raised to a power. It's often written as \( (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k}b^k \). Here, \( \binom{n}{k} \) represents the combination of \( n \) items taken \( k \) at a time.

In our exercise, we used the binomial theorem to understand how many combinations of objects can be formed. When choosing 10 objects out of a total of 31, the expansion \( (1+1)^{21} = 2^{21} \) shows all the ways to select subsets from 21 distinct objects. By summing combinations from 0 to 10, we essentially calculate part of this binomial expansion.

The importance of the binomial theorem here is that it provides a shortcut to calculate complex expansions often encountered in combinatorial problems.

Combinations

Combinations are all about choosing items from a collection, where the order of the chosen items does not matter. They are especially useful in problems where we need to count the possible selections of items from a larger set.

• To find the number of combinations, we use the formula \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \), where \( n \) is the total number of items to choose from, and \( r \) is the number of items to choose.
• In our context, when we decide how many distinct objects we should select, combinations help us calculate all these possibilities efficiently.

The goal is to choose enough distinct objects so that, together with the identical ones, the total number is 10. Combinatorics simplifies the counting process significantly, helping to analyze and manage vast possibilities.

Identical and Distinct Objects

One interesting aspect of combinatorics is handling identical and distinct objects, a recurring theme in many problems.

• "Identical objects" means the objects cannot be distinguished from one another. When you add identical items to a selection, it doesn't change the internal structure in terms of identity, only count.
• "Distinct objects" are unique and can be identified separately, making their selection non-trivial as each choice is potentially different.

In the exercise, we have 10 identical objects and 21 distinct ones. The challenge is deciding how many distinct objects to choose so the rest can be filled with identical ones to reach exactly 10. By focusing on the distinct selections first and then automatically knowing how many identical ones are needed, problems with complex mixed-object selections become more manageable and systematic.