Problem 53 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Consider a class of 5 girls and 7 boys. The number of different teams consisting of 2 girls and 3 boys that can be formed from this class, if there are two specific boys A and \(\mathrm{B}\), who refuse to be the members of the same team, is: (a) 500 (b) 200 (c) 300 (d) 350

Step-by-Step Solution

Answer

The correct answer is (c) 300.

Step 1: Understand the problem

We have a class of 5 girls and 7 boys. We need to form a team of 2 girls and 3 boys, where boys A and B cannot be on the same team. We will calculate the total possible teams and subtract the teams where A and B are together.

Step 2: Calculate total combinations of team members

First, calculate how many ways we can choose 2 girls from 5. This is done using combinations: \( \binom{5}{2} \). For 3 boys from 7, this is \( \binom{7}{3} \). The total number of teams can therefore be calculated as: \( \binom{5}{2} \times \binom{7}{3} \).

Step 3: Calculate comb. when A and B are together

If A and B are together, we need to choose 1 more boy from the remaining 5 boys (since B and A are already chosen, only 5 boys are left). This can be calculated as \( \binom{5}{1} \). We still choose 2 girls from 5, which remains \( \binom{5}{2} \). The number of such teams is \( \binom{5}{2} \times \binom{5}{1} \).

Step 4: Calculate the number of valid teams

Subtract the number of teams where A and B are together from the total number of teams. Total minus invalid gives valid: \((\binom{5}{2} \times \binom{7}{3}) - (\binom{5}{2} \times \binom{5}{1})\).

Step 5: Perform the calculations

Calculate the numerical values: \( \binom{5}{2} = 10 \), \( \binom{7}{3} = 35 \), \( \binom{5}{1} = 5 \). Therefore, the total number of teams is \(10 \times 35 = 350\). The number of teams invalid (A and B together) is \(10 \times 5 = 50\). Thus, valid teams are \(350 - 50 = 300\).

Step 6: Conclude with the correct answer

Based on our calculations, the number of valid teams that can be formed is 300.

Key Concepts

Combination formula

The combination formula is a fundamental concept in combinatorics, which helps us find the number of ways to choose items from a larger set, without considering the order of selection. For students, it's crucial to understand that the order does not matter in combinations, unlike permutations. This formula is denoted as \( \binom{n}{r} \), where \( n \) represents the total number of items to choose from, and \( r \) is the number of items to be selected.
For example, when calculating how many ways we can pick 2 girls out of 5, we use the combination formula: \( \binom{5}{2} \). The value is calculated as:

• \( \binom{5}{2} = \frac{5!}{2!(5-2)!} = 10 \)

Here, \(!\) stands for factorial, which means to multiply a series of descending natural numbers. Understanding this concept is pivotal in solving problems where group formation without regard to the sequence is needed.

Problem-solving strategy

A robust problem-solving strategy is key to tackling combinatorics exercises effectively. The first step is to always break down the problem clearly to understand the objective. In our example, we had distinct conditions: forming a team of 2 girls and 3 boys, with a condition that two specific boys cannot be together. Recognize all constraints, like the one involving boys A and B, to make informed calculations.
The strategy involves:

• Calculating total possible options first, disregarding the restriction.
• Then determine the outcomes that violate the conditions. This involves envisioning scenarios where the constraints are not met.
• Finally, subtract unwanted combinations from the total to find valid outcomes.

Having a clear strategy amplifies accuracy and ensures that no steps are missed in the calculation process.

Team formation

Understanding team formation in combinatorics is crucial as it involves selecting individuals or items to create subgroups. It becomes challenging with additional constraints like refusing members or required skills. For the exercise, forming a team from a group of students involves selecting a subset of students (2 girls and 3 boys) from a larger pool. The constraint that boys A and B cannot be in the same team adds an interesting challenge because it affects the typical combination calculations.
By applying combinatorics:

• You first calculate all possible teams by selecting without constraints.
• Then, adjust for specific conditions, like excluding combinations that do not fit the problem rules.

These exercises sharpen logic and mathematical thinking by requiring both creativity and precision in forming valid groups.

Arithmetic operations

Arithmetic operations are integral to performing calculations in combination problems. These operations include basic addition, subtraction, multiplication, and division, which are used here with factorials to determine combinations. In our problem, once combination values are determined using the formula \( \binom{n}{r} \), arithmetic operations are applied to assemble the final answer. We first calculate the total combinations using multiplication:

• Total combinations: \(10 \times 35 = 350\)

Then, subtraction is used to remove invalid outcomes:

• Invalid combinations: \(10 \times 5 = 50\)
• Valid teams: \(350 - 50 = 300\)

These calculations illustrate the high dependence on accurate arithmetic operations, ensuring the solutions result in correct conclusions.