Problem 142 - A Complete Resource Book in Mathematics for JEE Main

Question

Assertion: \(A\) set \(X\) contains \(n\) elements. Two subsets \(A\) and \(B\) of \(X\) are chosen at random. The probability that \(A\) and \(B\) have same number of elements is \(\frac{1 \cdot 3 \cdot 5 \ldots(2 n-1)}{2^{n}(n !)}\)

Step-by-Step Solution

Answer

The probability is \(\frac{1 \cdot 3 \cdot 5 \ldots (2n-1)}{2^n (n!)}\).

Step 1: Understand the Problem

We are given a set \(X\) with \(n\) elements. We need to determine the probability that two randomly chosen subsets \(A\) and \(B\) of \(X\) contain the same number of elements.

Step 2: Total Number of Subsets

For a set with \(n\) elements, the total number of subsets is \(2^n\). This is because each element can either be included or not included in a subset.

Step 3: Choosing Subsets with k Elements

For each subset, we can choose \(k\) elements from \(n\) elements. This can be done in \(\binom{n}{k}\) ways. So, each subset \(A\) or \(B\) having \(k\) elements can be done in \(\binom{n}{k}\).

Step 4: Equal Number of Elements in A and B

For any \(k\), both subsets \(A\) and \(B\) need to have \(k\) elements. The number of ways to select \(A\) is \(\binom{n}{k}\), and independently, the number of ways to select \(B\) is also \(\binom{n}{k}\).

Step 5: Sum Over All Possible k

The probability requires us to sum over all possible values of \(k\) from 0 to \(n\) for which \(A\) and \(B\) both have \(k\) elements. Therefore, the expression becomes \(\sum_{k=0}^{n} \left( \binom{n}{k} \right)^2\).

Step 6: Calculate the Probability

The probability that \(A\) and \(B\) have the same number of elements is the ratio of favorable outcomes to total outcomes. As every subset of \(X\) has \(2^n\) possibilities independently, the total ways of selecting both \(A\) and \(B\) are \((2^n)^2 = 4^n\).

Step 7: Use Known Combinatorial Identity

We apply the identity \(\sum_{k=0}^{n} \binom{n}{k}^2 = \binom{2n}{n}\). This simplifies the numerator of our probability formula.

Step 8: Final Probability Calculation

Substituting the identity into our problem, the probability becomes \(\frac{\binom{2n}{n}}{4^n}\). Since \(\binom{2n}{n} = \frac{(2n)!}{(n!)^2}\), we equate it with the given form, \(\frac{1 \cdot 3 \cdot 5 \ldots (2n-1)}{2^n n!}\).

Step 9: Verify Identity for Result

The expression \(\binom{2n}{n} = \frac{(2n)!}{(n!)^2}\) simplifies using double factorial and the given result \(\frac{1 \cdot 3 \cdot 5 \ldots (2n-1)}{2^n n!}\), confirming that both expressions are equivalent forms using combinatorial properties.

Key Concepts

Combinatorial Identity

In probability theory, combinatorial identities are equations that involve combinatorial quantities, like binomial coefficients and factorials. These identities help us simplify complicated expressions by discerning underlying mathematical relationships.

One well-known combinatorial identity is the equality \[\sum_{k=0}^{n} \binom{n}{k}^2 = \binom{2n}{n}.\]This identity plays a crucial role in calculating probabilities involving subsets. It tells us that if we sum the squares of the binomial coefficients \(\binom{n}{k}\) from 0 to \(n\), the result equals \(\binom{2n}{n}\).

This identity is useful for determining probabilities of events where the same number of items need to be selected from two groups, as done in our exercise. Using such identities allows for elegant solutions that might not be apparent with straightforward combinatorial arguments.

Binomial Coefficient

The binomial coefficient, denoted as \(\binom{n}{k}\), represents the number of ways to choose \(k\) elements from a set of \(n\) elements, regardless of the order of selection. The formula for calculating a binomial coefficient is given by:\[\binom{n}{k} = \frac{n!}{k!(n-k)!}.\]These coefficients are fundamental to probability calculations involving combinations and permutations.

In the context of the exercise, binomial coefficients determine the number of subsets \(A\) and \(B\) that have an equal number of elements. Specifically, the expression \(\binom{n}{k}^2\) represents picking \(k\) elements for subset \(A\) and \(k\) elements for subset \(B\) independently.

• The factorial \(n!\) denotes the product of all positive integers up to \(n\). It reflects the total number of ways to arrange \(n\) items.
• Important properties of binomial coefficients include symmetry \(\binom{n}{k} = \binom{n}{n-k}\), and Pascal's identity \(\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}\).

These coefficients, together with various identities, allow for simplification of many probability problems by providing a compact way to express combinations.

Subsets and Elements

Subsets are smaller sets created from selecting elements of a larger set without regard to the order of elements. The original set \(X\) in our problem has \(n\) elements, and for each element of \(X\), there are two possibilities: it is either included in a subset or not.

This leads to the total number of subsets in a set \(X\) containing \(n\) elements being \(2^n\). Subsets can vary in size, from the empty set containing no elements to the entire set \(X\) itself.

In probability theory, understanding subsets is essential because:

• The selection process can yield different possible outcomes based on the size of subsets.
• Set theory and combinatorial methods help quantify these subsets, essential for calculating probabilities and validating identities.

In our exercise, the emphasis almost exclusively concerns subsets of equal size. We seek the likelihood that subsets \(A\) and \(B\) drawn from the set \(X\) have the same number of elements. This involves assessing all such combinations of elements and the probability distribution of selecting a pair of equally sized subsets. By understanding subsets and their formation, one gains valuable insights into the combinatorial configurations required for complex probabilistic evaluations.