Problem 68 - Algebra for JEE ADVANCED

Question

The number of ways of choosing a committee of two women and three men from five women and six men, if \(\mathrm{Mr}\). A refuses to serve on the committee if \(\mathrm{Mr}\). \(\mathrm{B}\) is a member and \(\mathrm{Mr}\). B can only serve, if Miss \(\mathrm{C}\) is the member of the committee is a. 60 b. 84 c. 124 d. none of these

Step-by-Step Solution

Answer

The answer is 124 (option c).

Step 1: Understanding the Problem

We need to form a committee of 2 women and 3 men from a group of 5 women and 6 men. Specific conditions are given: Mr. A refuses to be on the committee if Mr. B is there, and Mr. B can only be there if Miss C is also on the committee. We need to find the number of valid ways to form such a committee.

Step 2: Total Ways Without Constraints

First, calculate the total number of ways to select 2 women and 3 men without any constraints.The number of ways to choose 2 women from 5 is \( \binom{5}{2} \), and for 3 men from 6 is \( \binom{6}{3} \). Calculate these values: \( \binom{5}{2} = 10 \) and \( \binom{6}{3} = 20 \). Thus, the total number of committees is \( 10 \times 20 = 200 \).

Step 3: Applying the Constraint Without Mr. B

Calculate the number of committees that do not include Mr. B. We only choose from the other 5 men.So, choosing 3 men from 5 is \( \binom{5}{3} \).Calculate this: \( \binom{5}{3} = 10 \).Multiply with the ways to choose 2 women from 5:\( 10 \times 10 = 100 \) ways where Mr. B is not on the committee.

Step 4: Applying the Constraint With Mr. B and Miss C

If Mr. B is in the committee, so must Miss C, which leaves us choosing 1 more woman out of the remaining 4, and 2 more men from the remaining 5.\( \binom{4}{1} = 4 \), \( \binom{5}{2} = 10 \).Multiply these: \( 4 \times 10 = 40 \) ways when Mr. B and Miss C are in the committee.

Step 5: Subtracting Invalid Committees

Out of the cases where Mr. B is present and valid, none should include Mr. A according to the constraints. Calculate those cases.Choose 2 men (excluding Mr. A and B) from the other 4:\( \binom{4}{2} = 6 \).So the valid cases are \( 4 \times 6 = 24 \).

Step 6: Final Calculation

Add the valid cases from steps without Mr. B and with Mr. B.\( 100 + 24 = 124 \).Thus, the number of valid ways to form the committee is 124.

Key Concepts

Combinatorial Problem-Solving

Combinatorial problem-solving refers to the mathematical technique used to determine the number of possible arrangements or selections in a given scenario. In problems involving combinatorics, we often deal with selecting items or forming groups under specific conditions. To solve such problems efficiently, one must identify the total number of possibilities and then apply any given constraints or conditions to narrow down to valid solutions.

• Start by assessing the entire problem in a clear manner.
• Identify the subsets of elements involved, such as sets of people or objects.
• Use mathematical tools and concepts like permutations and combinations to determine the number of ways to perform the selections.

Breaking down the problem into sequential steps makes it easier to incorporate any conditions (like Mr. A's or Mr. B's preferences) and calculate the exact solution.

Binomial Coefficient

The binomial coefficient is a handy tool used to find the number of ways to choose a subset from a larger set, without regard to the order of selection. It is denoted by \( \binom{n}{k} \), which reads "n choose k," representing the number of ways to choose \( k \) items from \( n \) items.In the context of our committee problem:

• The formula is \( \binom{n}{k} = \frac{n!}{k! (n-k)!} \), where "!" signifies factorial, the product of all positive integers up to a certain number.
• Choosing 2 women from 5 involves calculating \( \binom{5}{2} \), which is simplified as \( \frac{5 \times 4}{2 \times 1} = 10 \).
• Similarly, for selecting 3 men from 6, you compute \( \binom{6}{3} = 20 \).

Exceptionally useful, the binomial coefficient allows us to compute large combinations without the need for extensive trial and error.

Committee Selection

Committee selection embodies a classic combinatorial task: choosing a smaller group meeting certain criteria from a larger one. In our example, a committee must be formed under the constraints that involve preferences and conditions regarding members Mr. A, Mr. B, and Miss C. To grasp committee selection effectively:

• Understand the different possible ways to form committee members, such as the number of combinations to select women and men.
• Account for the restrictions like Mr. A refusing to serve if Mr. B is present.
• Engage the given conditions like Mr. B needing Miss C's presence on the committee to include or exclude different selections.
• Calculate these separately and then aggregate valid combinations ensuring no double-counting.

This structured approach simplifies complex selection tasks and ensures a correct count.

Conditional Probability

Conditional probability in combinatorics focuses on determining the likelihood or count of an event happening given certain pre-existing conditions. In our example about forming a committee, the conditional nature revolves around obligatory or prohibitive pairings like Mr. A's and Mr. B's preferences. Steps to manage conditional scenarios effectively:

• Identify independent events that do not influence one another and dependent events like Mr. B's condition requiring Miss C.
• Modify your formula or approach to factor in these specific conditions rather than treating the elements as isolated choices.
• Example: If Mr. B must have Miss C present, calculate with both included, then evaluate if further conditions (like excluding Mr. A) influence these selections.

Mastering conditional probability not only improves combinatorial problem-solving but also enhances logical reasoning and ensures accurate results by methodically addressing conditions.