Problem 2 - Algebra for JEE ADVANCED

Question

Kanchan has 10 friends among whom two are married to each other. She wishes to invite five of them for a party. If the married couples refuse to attend separately, then the number of different ways in which she can invite five friends is a. \({ }^{8} C_{5}\) b. \(2 \times{ }^{8} C_{3}\) c. \({ }^{10} C_{5}-2 \times{ }^{8} C_{4}\) d. none of these

Step-by-Step Solution

Answer

Option b, 112 ways.

Step 1: Understanding the Problem

Kanchan has 10 friends. Among them, there's a married couple who refuse to attend separately. She needs to invite five friends to her party. Our goal is to count the number of ways to select 5 friends under these conditions.

Step 2: Consider the Married Couple Together

Since the married couple either attends together or not at all, we treat them as a single entity or 'block'. This reduces the total number of entities from 10 to 9 (8 individual friends + 1 couple).

Step 3: Counting Cases with the Couple

If the couple is invited, we need to choose 3 more friends from the remaining 8 friends: \[ \binom{8}{3} \] Thus, the number of ways to invite 3 additional friends from 8 is 56.

Step 4: Final Calculation Considering Both Choices of the Couple

Since the couple can either attend together or not at all, we consider only the case where they attend. The expression becomes: \[ 2 \times \binom{8}{3} = 2 \times 56 = 112 \]

Step 5: Conclusion

The correct option, based on our calculations, is b. \(2 \times \binom{8}{3}\).

Key Concepts

Binomial Coefficient

In the world of combinatorics, the binomial coefficient is a fundamental concept that helps us calculate how many ways we can choose a certain number of items from a larger set. This is usually expressed using the notation \( \binom{n}{k} \), which stands for the number of combinations of \( n \) items taken \( k \) at a time. The formula to compute this is:\[\binom{n}{k} = \frac{n!}{k!(n-k)!}\]where \( n! \) (called "n factorial") is the product of all positive integers up to \( n \). For example, \( \binom{8}{3} \) equals 56 because there are 56 ways to select 3 friends out of 8.
This concept is crucial when addressing questions involving grouping and selection, as seen in the exercise where Kanchan selects friends for a party. Understanding how to apply the binomial coefficient effectively is key to solving such problems efficiently.

Permutation and Combination

Permutations and combinations are two different ways of arranging items. While permutations are concerned with the order of items, combinations are not. In our exercise, we are interested in combinations since Kanchan wants to select friends without worrying about the order in which they are invited.

• **Permutations**: Used when order matters. Calculated as \( n! / (n-k)! \).
• **Combinations**: Used when order does not matter. Calculated using binomial coefficients, as \( \binom{n}{k} \).

In Kanchan's scenario, permutations are irrelevant because a friend group for a party is the same regardless of the order in which they are invited. The couple and other friends are simply selected, illustrating the concept of combinations.

Problem Solving

Effectively approaching problem-solving requires a strategic breakdown of the conditions and constraints provided. In Kanchan's problem, the main challenge is the married couple's condition to attend together or not at all. Here's a simplified approach to solve such problems:
1. **Identify Constraints**: Recognize special conditions, such as the couple's need to attend together. 2. **Break Down the Problem**: Treat the couple as a single entity to simplify calculations without breaching the problem's constraints. 3. **Calculate Using Binomial Coefficients**: Use the binomial coefficient to determine the number of ways to select the other required friends, which simplifies the process significantly. 4. **Consider Variations**: Since the couple has two "modes" (attending or not), account for both in your calculations.
Combining these steps sharpens your ability to tackle complex combinatorial problems, just like navigating the nuances of inviting the perfect party group with level-headed calculations.