Question

If 6 objects are distributed at random among 6 persons, the probability that atleast one of them will not get any thing is : (a) 6.(6!) (b) \(\frac{5^{6}}{6 !}\) (c) \(\frac{6^{6}-6 !}{6^{6}}\) (d) none of these

Short answer

Answer: (c) \(\frac{6^{6}-6 !}{6^{6}}\)

Step-by-step solution

Find the total number of ways to distribute the objects

Using the total number of ways to distribute n objects among n persons formula, we can say that the objects can be distributed in: \(6^6\) ways.

Calculate the number of ways of excluding one person

We can choose any one person to not receive an object. After that, we can distribute the 6 objects among the remaining 5 persons in the following ways: \(C(6,1) \cdot 5^6\) ways.

Calculate the number of ways of excluding two persons

We can choose any two persons out of the 6 persons who will not receive an object, we can distribute the 6 objects among the remaining 4 persons in the following ways: \(C(6,2) \cdot 4^6\) ways.

Calculate the number of ways of excluding three persons

Following the same logic, if we choose to exclude three persons, we can distribute the objects in the following ways: \(C(6,3) \cdot 3^6\) ways.

Calculate the number of ways of excluding four persons

If we choose to exclude four persons, we can distribute the objects in the following ways: \(C(6,4) \cdot 2^6\) ways.

Calculate the number of ways of excluding five persons

Excluding five persons would mean giving all the 6 objects to a single person. That can be done in the following ways: \(C(6,1)\) ways.

Apply the Inclusion-Exclusion Principle

Sum up the number of ways found in steps 2 to 6 and subtract the even-numbered step counts from the odd-numbered step counts to find the total number of ways of distributing so that at least one person is excluded. Total number of ways to distribute = \(C(6,1) \cdot 5^6 - C(6,2) \cdot 4^6 + C(6,3) \cdot 3^6 - C(6,4) \cdot 2^6 + C(6,1)\)

Calculate the probability

Divide the total number of ways from step 7 by the total number of ways found in step 1: \(P = \frac{C(6,1) \cdot 5^6 - C(6,2) \cdot 4^6 + C(6,3) \cdot 3^6 - C(6,4) \cdot 2^6 + C(6,1)}{6^6}\)

Simplify the expression

After simplifying the expression from step 8, we arrive at the probability: \(P = \frac{6^6 - 6!}{6^6}\) So the correct answer is (c) \(\frac{6^{6}-6 !}{6^{6}}\).

Key concepts

Inclusion-Exclusion Principle

In probability theory and combinatorics, the Inclusion-Exclusion Principle is a powerful tool used to calculate the cardinality of the union of multiple sets. It helps account for overlapping elements among the sets, ensuring an accurate total. When distributing objects or dealing with events, some might overlap, leading to double-counting if not addressed properly.
To adjust for overlaps, the principle involves:

• Adding the sizes of the individual sets.
• Subtracting the sizes of the pairwise intersections, since they are counted twice in the first step.
• Adding back the sizes of the intersection of three sets, since each three-set intersection had been subtracted too many times.
• Continue this alternating process for intersections of more sets.

In the provided exercise, the Inclusion-Exclusion Principle helps to calculate the number of ways to distribute objects while excluding certain persons. When applied, this technique subtracts and adds the number of ways each subset is included, ensuring each configuration is only counted once. It's especially useful when dealing with complex problems involving overlapping conditions, like making sure that each person receives at least one object.

combinatorics

Combinatorics is the branch of mathematics dealing with combinations, permutations, and counting, which are foundational for probability theory. It explores how to count, arrange, and analyze discrete structures. In the context of the problem, combinatorics helps to determine the number of ways objects can be distributed among people.
For instance, consider distributing 6 objects among 6 persons without any being left out. There are various constraints: each must receive something, or some may receive nothing.
The exercise leverages combinatorial calculations:

• Ways to distribute objects to persons: Using formulas like \(n^n\), which signifies each of the \(n\) persons can receive each of \(n\) objects.
• Exclusions: Calculating combinations of persons who might receive no objects, expressed as \(C(n,k)\), where \(C\) is the combination operator, \(n\) is the total, and \(k\) is those chosen for exclusion.

Mastering these combinatorial concepts enables a deeper understanding of how particular distributions are constructed and confirms the calculations' accuracy.

discrete mathematics

Discrete mathematics encompasses various topics, including the study of countable, distinct objects. It's essential for understanding probability theory and combinatorics. Key areas in discrete mathematics include set theory, graph theory, and number theory, among others.
In this problem, discrete mathematics is at work when analyzing the allocation of objects to people. The task is formed on foundations of finite sets and operations on these sets. Set theory, a component of discrete mathematics, informs how we calculate distributions and their exclusions using the Inclusion-Exclusion Principle.
Engagement with discrete structures means:

• Thinking in terms of specific, separate objects, rather than continuous systems.
• Utilizing logical structures like combinations and permutations for problem-solving.
• Applying mathematical reasoning to finite scenarios, such as determining probabilities in configurations.

Getting comfortable with these discrete math principles helps when tackling real-world problems involving distinct items, events, or cases where order and arrangement matter.