Problem 80 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If the number of ways in which \(n\) different things can be distributed among \(n\) persons so that at least one person does not get any thing is 232 . Then, \(n\) is equal to (A) 3 (B) 4 (C) 5 (D) None of these

Step-by-Step Solution

Answer

The correct solution cannot be determined from the given choices based on standard logic.

Step 1: Understanding the Surface of the Problem

We need to find the number \( n \) such that when \( n \) different things are distributed among \( n \) people, at least one person gets nothing. The result is given as 232. This means we need to use the concept of derangements, as it ensures that no person gets the thing with the corresponding number.

Step 2: Calculating Total Ways without Restrictions

First, calculate the total number of ways to distribute \( n \) items among \( n \) people if there were no restrictions. This is simply \( n! \) (factorial of \( n \)).

Step 3: Applying Inclusion-Exclusion Principle

According to the inclusion-exclusion principle, we subtract the number of ways where at least one person gets nothing. This is calculated by \( n! - D(n) = 232 \), where \( D(n) \) is the number of derangements of \( n \) items.

Step 4: Finding the Derangement Formula

Derangement \( D(n) \) is calculated as: \[ D(n) = n! \left( 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \cdots + \frac{(-1)^n}{n!} \right) \]

Step 5: Testing Potential Values for n

Test values for \( n = 3, 4, 5 \) to see which satisfies the equation \( n! - D(n) = 232 \). Calculate \( D(n) \) using the given formula and check.

Step 6: Verification for n=5

For \( n=5 \): - Calculate \( 5! = 120 \). - Calculate \( D(5) = 5! \left( 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} \right) = 44 \). - Hence, \( 5! - D(5) = 120 - 44 = 76 \), not matching 232.

Step 7: Verification for n=4

For \( n=4 \): - Calculate \( 4! = 24 \). - Calculate \( D(4) = 4! \left( 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} \right) = 9 \). - Hence, \( 4! - D(4) = 24 - 9 = 15 \), not matching 232.

Step 8: Verification for n=6

For \( n=6 \): - Calculate \( 6! = 720 \). - Calculate \( D(6) = 6! \left( 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} + \frac{1}{6!} \right) = 265 \). - Verify: The equation \( 720 - 265 = 455 \), does not match 232.

Step 9: Conclusion and Answer

The correct method involves further calculations, checking each possibility in detail. The problem statement itself could be inconsistent, as testing common values led to no correct answer based on this mathematical framework.

Key Concepts

Derangements

A derangement is a fascinating concept in permutation problems. It refers to a permutation of a set where none of the elements appear in their original positions. Imagine you have a set of items, each assigned a number, and you want to rearrange them so that no item is in the same position as before.

This principle is essential in our exercise problem, where we want to distribute items among people such that no one gets their intended corresponding item. Derangements help us calculate the number of ways this can happen efficiently.

• A standard derangement is calculated using the formula: \[ D(n) = n! \left( 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \cdots + \frac{(-1)^n}{n!} \right) \]
• This formula involves alternating sums and differences of factorial inverses, making it robust for handling derangement problems of varying sizes.

Factorials

Factorials, denoted by an exclamation mark (!), are a key concept in permutations and combinations. The factorial of a positive integer \( n \), represented as \( n! \), is the product of all positive integers up to \( n \).

Understanding factorials is crucial for calculating the total number of ways to arrange \( n \) items. This has direct implications in our problem where we determine the possible distributions of items with and without the constraints imposed by derangements.

• For instance, \( 5! \) is calculated as \( 5 \times 4 \times 3 \times 2 \times 1 = 120 \).
• Factorials grow rapidly, which can be both useful for large permutations and complex in computations without proper simplifications like those offered by derangements.

Inclusion-Exclusion Principle

The inclusion-exclusion principle is a combinatorial method used to count the number of elements in the union of several sets. It effectively helps us refine our counting by considering overlaps, ensuring we don't count elements more than once.

In the context of our exercise, this principle helps determine how many ways we can distribute items such that at least one person does not receive anything. We start with the total number of possible distributions and subtract the cases where everyone receives something, which directly relates to derangements.

• The formula we used is: \( n! - D(n) = \text{number of restricted arrangements} \).
• This shows how inclusion-exclusion allows correction from overcounting by considering constraints like avoiding intended assignments.