Problem 69 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If the number of 5 -element subsets of the set \(A=\left\\{a_{1}, a_{2}, \ldots, a_{20}\right\\}\) of 20 distinct elements is \(k\) times the number of 5 -element subsets containing \(a_{4}\), then \(k\) is (a) 5 (b) \(\frac{20}{7}\) (c) 4 (d) \(\frac{10}{3}\)

Step-by-Step Solution

Answer

The value of \(k\) is 4 (c).

Step 1: Identify the Total Number of 5-element Subsets

The total number of subsets of 5 elements that can be formed from a set of 20 distinct elements is given by the combination formula \( \binom{20}{5} \). Calculate it using \( \binom{20}{5} = \frac{20!}{5!(20-5)!} \).

Step 2: Calculate the Subsets Containing Specific Element

For subsets containing a specific element, like \(a_4\), we first include \(a_4\) in the subset, which means we need to choose the remaining 4 elements from the other 19 elements. This is given by \( \binom{19}{4} \).

Step 3: Establish the Equation

According to the question, the total number of 5-element subsets without any condition is \(k\) times the number of 5-element subsets containing \(a_4\). We express this as: \( \binom{20}{5} = k \times \binom{19}{4} \).

Step 4: Solve for \(k\)

Substitute the values calculated: \( \binom{20}{5} = \frac{20!}{5!15!} = 15504 \) and \( \binom{19}{4} = \frac{19!}{4!15!} = 3876 \). Now solve for \(k\): \( 15504 = k \times 3876 \). Thus, \( k = \frac{15504}{3876} \).

Step 5: Simplify to Calculate \(k\)

Simplify the expression: \( \frac{15504}{3876} = 4 \). Therefore, the value of \(k\) is 4.

Key Concepts

Binomial coefficient

The binomial coefficient, often denoted as \( \binom{n}{k} \), represents the number of ways to choose \( k \) items from a set of \( n \) distinct items without regard to the order of selection. It is a fundamental concept in combinatorics, which is the branch of mathematics dealing with combinations, permutations, and counting. You can calculate the binomial coefficient using the formula: \[\binom{n}{k} = \frac{n!}{k!(n-k)!} \]where \( n! \) (read as "n factorial") is the product of all positive integers up to \( n \). For instance, to compute \( \binom{20}{5} \), like in the exercise, you would arrange 20 items into 5 slots, resulting in \[\binom{20}{5} = \frac{20!}{5!(20-5)!} = 15504.\]The binomial coefficient tells us that there are 15,504 possible combinations when selecting 5 items from 20.

Subset selection

Subset selection involves choosing a number of elements from a larger set. In the context of combinatorics, this process ensures that all possible combinations of elements are considered, without accounting for any specific order among them. In the given exercise, a subset selection process is used to form subsets of 5 elements, which include and exclude a specific element \(a_4\). To find the number of all 5-element subsets, we calculated \( \binom{20}{5} \). However, when \(a_4\) must be included, the problem reduces to finding the number of combinations of the remaining elements:- First, include \(a_4\) in the subset.- Then, choose 4 additional elements from the other 19 available.- Calculate \( \binom{19}{4} \) to find the exact number of possible subsets containing \(a_4\), which equals 3,876.Subset selection helps in systematically exploring all potential groupings within a set.

Combination formula

The combination formula, represented as \( \binom{n}{k} \), is the crucial tool for solving problems involving selection without replacement. The formula helps determine how many ways you can select \( k \) items from a set of \( n \) distinct items where the order doesn't matter.Here's how it works, step-by-step:

• Determine the total number of items \( n \).
• Decide the number \( k \) of items to choose.
• Use the formula: \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \).

In our scenario, we calculated \( \binom{19}{4} \) for selecting 4 items from a set where one was predetermined (\(a_4\)), resulting in 3,876 unique combinations.The combination formula is pivotal because it simplifies complex counting problems into manageable computations, effectively answering questions about grouping and arrangements.