Problem 9 - Algebra 2 Course in Mathematics for the IIT-JEE and Other Engineering Exams

Question

Five persons entered the lift cabin on the ground floor of an 8 -floor house. Suppose that each of them independently and with equal probability can leave the cabin at any floor beginning with the first. Find out the probability of all 5 persons leaving at different floors. (a) \(\frac{1}{7^{5}}\) (b) \(\frac{1}{{ }^{7} P_{5}}\) (c) \(\frac{{ }^{7} P_{5}}{7^{5}}\) (d) None of these

Step-by-Step Solution

Answer

The probability is \(\frac{{ }^{7}P_{5}}{7^{5}}\), which is option (c).

Step 1: Understand the Problem

We need to calculate the probability that each of the 5 people leaves the lift on a different floor in an 8-floor house. Floors can range from 1 to 7 as they entered on the ground floor.

Step 2: Determine Total Possible Outcomes

Each person can leave the lift at any one of 7 floors. Therefore, the total number of ways the 5 people can leave is given by \[7^5\]This is because each person has 7 choices.

Step 3: Determine Favorable Outcomes

To have all 5 people exit on different floors, choose 5 floors out of 7 for them to exit from. This can be done in \(\binom{7}{5}\) ways. Once the floors are selected, arrange 5 people in the 5 different chosen floors, i.e. \[5!\] ways. The number of favorable ways is \[\binom{7}{5} \times 5!\].

Step 4: Calculate the Number of Favorable Outcomes

Calculate the number of favorable outcomes using permutations.\[\binom{7}{5} = \frac{7!}{5!(7-5)!} = \binom{7}{2}\]\[\binom{7}{2} = 21\]Also, calculate the arrangements:\[5! = 120\]Thus, the number of favorable outcomes is \[21 \times 120 = 2520\].

Step 5: Calculate the Probability

The probability that all 5 persons leave at different floors is the ratio of the number of favorable outcomes to the total possible outcomes:\[\frac{\text{Number of favorable outcomes}}{\text{Total possible outcomes}} = \frac{2520}{7^5}\]Calculate:\[\frac{2520}{16807} = \frac{2520}{16807} = \frac{\binom{7}{5} \times 5!}{7^5}\]This simplifies to the probability: \[\frac{^{7}P_5}{7^5}\].

Step 6: Match with Options

From the given options, option (c) \(\frac{{ }^{7}P_{5}}{7^{5}}\) matches with our calculated probability.

Key Concepts

Permutations

Permutations are an essential concept in probability and combinatorics, often used to determine how many ways a set of items can be arranged. In the context of the elevator problem, permutations help us figure out how to arrange 5 people on 5 different floors among the available options.

When thinking about permutations, it’s important to understand that order matters. For example, if you have 5 people and you want them to leave on different floors, it matters which person leaves on which floor. This is why we use the factorial notation, denoted as \(n!\). It represents the product of all positive integers up to \(n\). So, \(5!\) means:

• 1 x 2 x 3 x 4 x 5 = 120

This tells us there are 120 different ways to order the 5 people.

Additionally, when choosing distinct floors for them to leave on, we employ permutations again. Selecting 5 floors among 7 can be calculated using a permutation formula, expressed as \(^7P_5\). This calculates the number of ways to arrange 5 people across 5 chosen floors, taking order into account. The correct permutation calculation derives from the formula:\[\frac{7!}{(7-5)!} = ^7P_5 = 2520\]

Combinatorics

Combinatorics involves counting, arranging, and grouping sets of elements, essential for solving various probability problems. In our elevator scenario, we used combinatorics to determine how many ways 5 people could leave the elevator on different floors. This is where understanding permutations, combinations, and their distinct applications become critical.

Combinations, which were indirectly used here, focus on selecting items without concern for order. The number of ways to select 5 floors out of 7 is calculated using combinations, expressed as \(\binom{7}{5}\). They count the possible groupings of floors irrespective of exit order:

• \(\binom{7}{5} = \frac{7!}{5!(7-5)!}\)
• \(\binom{7}{5} = 21\)

Once we have chosen floors, we then use permutations to arrange the people onto those floors, considering the order they exit. This showcases how combinatorics interweaves combinations and permutations to resolve complex problems efficiently.

Discrete Mathematics

Discrete mathematics focuses on structures that are distinct or separable, like integers, graphs, or in this case, floors in a building. It's crucial in a wide variety of applications, particularly those involving counting and structure. In our problem, floors are discrete units, and we need to address them individually.

The finite set of floors where people can exit the elevator forms a classic example of how discrete mathematics functions. Unlike continuous mathematics where solutions are often found in real numbers, discrete mathematics deals with finite possibilities. This is apparent in our calculation where the floors are integer values ranging from 1 to 7, inclusive. Each floor represents a precise, separate choice, not a point within an interval.

More specifically, applying our probability calculation involves counting techniques which are a discrete concept: the task is to count how many ways individuals can exit the lift while adhering to the rules (i.e., exiting at different floors). This showcases the power of discrete mathematics in providing exact solutions for specific, finite systems, as opposed to approximations or areas.