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

Question

An elevator starts with \(m\) passengers and stops at \(n\) floors \((m \leq n)\). The probability that no two passengers alight at the same floor is (A) \(\frac{n_{m}}{m^{n}}\) (B) \(\frac{{ }^{n} P_{\text {w }}}{n^{m}}\) (C) \(\frac{n_{C_{m}}}{m^{n}}\) (D) \(\frac{n^{n} C_{m}}{n^{m}}\)

Step-by-Step Solution

Answer

The probability is given by option (B): \(\frac{{ }^{n} P_{m}}{n^{m}}\).

Step 1: Understand the Problem

We need to calculate the probability that no two passengers alight from the elevator on the same floor. This implies that every passenger chooses a different floor to alight.

Step 2: Calculate Total Ways Passengers Can Alight

Each of the \(m\) passengers can alight on any of the \(n\) floors. Thus, the total number of unrestricted possibilities for passenger alightments is \(n^m\).

Step 3: Determine the Favorable Outcomes

If no two passengers can alight on the same floor, this is a permutation of \(m\) passengers over \(n\) different floors. There are \(P(n, m) = \frac{n!}{(n-m)!}\) favorable ways.

Step 4: Compute the Desired Probability

The probability is the ratio of favorable outcomes to all possible outcomes: \[ P = \frac{P(n, m)}{n^m} = \frac{n!}{(n-m)! \cdot n^m} \]

Step 5: Match the Result with Given Options

The probability expression \(\frac{P(n, m)}{n^m} = \frac{{ }^{n} P_{m}}{n^{m}}\) matches option (B).

Key Concepts

Permutations

Permutations are fundamental in understanding problems that require arranging or selecting items where order matters. In our exercise, the concept of permutations arises when determining how passengers alight on different floors without any two alighting on the same floor.
In simpler terms, a permutation is a unique order arrangement of a subset of items from a larger set. For example, if we have 3 passengers and 5 floors, a permutation problem asks us to find how many different ways 3 passengers can alight on 3 distinct floors.
The formula for permutations when choosing a subset of items, say finding the number of ways to arrange \(m\) items out of \(n\), is \(P(n, m) = \frac{n!}{(n-m)!}\). This reflects the numerous ways to pick and order \(m\) passengers from \(n\) floors without repetition.

Combinatorial Probability

Combinatorial probability combines counting techniques, such as permutations and combinations, to calculate the likelihood of specific events.
In the context of our exercise, the probability that no two passengers alight on the same floor is such a problem.
The approach involves comparing the number of favorable outcomes (arrangements where passengers alight on different floors) to the total possible outcomes (all possible ways passengers could alight).

• Total possible outcomes: Each of the \(m\) passengers can choose any of the \(n\) floors, resulting in \(n^m\) combinations.
• Favorable outcomes: Since no two passengers should alight on the same floor, it involves permutations, given as \(P(n, m)\).

The probability is then determined by the formula: \[ P = \frac{P(n, m)}{n^m} \]This comparison yields the probability of passengers alighting on separate floors, solving our exercise effectively.

Passenger Alighting Problem

The passenger alighting problem is a real-life scenario exemplified in combinatorics.
It involves calculating the probability that passengers in an elevator choose distinct floors to alight on.
The challenge arises from needing to ensure each passenger selects a unique floor among many possible destinations.
To solve the problem, we:

• Identify that the event requires a permutation, preventing any two passengers from alighting on the same floor.
• Calculate the total unrestricted possibilities for alightment, which is \(n^m\), where \(n\) is the number of floors and \(m\) is the number of passengers.
• Use the permutation formula \(P(n, m)\) to determine how these passengers can alight at distinct floors.

The final step involves computing the probability using these calculations, helping us understand the likelihood of such events in the real world, and illustrates the practical application of combinatorial probability.