The number of ways in which 3-holiday tickets can be given to 20 employees of an organization if each employee is eligible for any one or more of the tickets, is
8000
Understanding the Problem: Distributing Holiday Tickets
The question asks for the number of ways to distribute 3 distinct holiday tickets among 20 employees. A key condition mentioned is that "each employee is eligible for any one or more of the tickets". This condition is crucial for determining the approach to solve this counting problem.
Let's break down what the condition "each employee is eligible for any one or more of the tickets" means:
- Each ticket is distinct (Ticket 1, Ticket 2, Ticket 3).
- For the first ticket, any of the 20 employees can receive it.
- For the second ticket, any of the 20 employees can receive it, regardless of who received the first ticket. The same employee can receive multiple tickets.
- For the third ticket, any of the 20 employees can receive it, regardless of who received the first two tickets.
This scenario is a case of permutations with repetition, where we are selecting from the set of employees (the recipients) for each of the tickets (the items being distributed). Since the order matters (Ticket 1 going to Employee A and Ticket 2 to Employee B is different from Ticket 1 going to Employee B and Ticket 2 to Employee A, and the tickets are distinct) and repetition is allowed (an employee can receive multiple tickets), we can determine the number of choices for each ticket independently.
Step-by-Step Calculation for Ticket Distribution
We have 3 holiday tickets to distribute among 20 employees. Let's consider the choices for each ticket:
- Ticket 1: This ticket can be given to any of the 20 employees. So, there are 20 possible choices for Ticket 1.
- Ticket 2: This ticket can also be given to any of the 20 employees, as employees can receive more than one ticket. So, there are 20 possible choices for Ticket 2.
- Ticket 3: Similarly, this ticket can be given to any of the 20 employees. So, there are 20 possible choices for Ticket 3.
Since the choice of employee for each ticket is independent of the choices for the other tickets, the total number of ways to distribute the 3 holiday tickets is the product of the number of choices for each ticket.
Total number of ways = (Choices for Ticket 1) × (Choices for Ticket 2) × (Choices for Ticket 3)
Total number of ways = $20 \times 20 \times 20$
Total number of ways = $20^3$
Calculating the Final Number of Ways
Now, we calculate the value of $20^3$:
$20^3 = 20 \times 20 \times 20 = 400 \times 20 = 8000$
Thus, there are 8000 different ways to give 3 holiday tickets to 20 employees when each employee is eligible for any one or more of the tickets.
Summary of Distribution Ways
Let's summarize the calculation:
| Ticket | Number of Employee Choices |
|---|---|
| Ticket 1 | 20 |
| Ticket 2 | 20 |
| Ticket 3 | 20 |
Total ways = $20 \times 20 \times 20 = 8000$.
Revision Table: Holiday Ticket Distribution
| Concept | Details |
|---|---|
| Problem Type | Distribution of distinct items (tickets) to distinct recipients (employees) with repetition allowed for recipients. |
| Items (n) | 3 distinct tickets |
| Recipients (r) | 20 distinct employees |
| Condition | Each employee eligible for one or more tickets (repetition allowed). |
| Formula Used | $r^n$ (Number of ways to distribute n distinct items into r distinct bins with repetition allowed per bin) |
| Calculation | $20^3 = 8000$ |
Additional Information: Combinatorics Concepts
This problem is an example of counting arrangements where repetition is permitted. It's helpful to compare this to other common combinatorics scenarios:
- Permutations without Repetition: Used when arranging distinct items where each item can be used only once. For example, arranging 3 specific employees in a line ($P(20, 3) = 20 \times 19 \times 18$).
- Combinations without Repetition: Used when selecting a group of distinct items where the order doesn't matter and each item can be used only once. For example, choosing 3 employees out of 20 to form a committee ($\binom{20}{3} = \frac{20 \times 19 \times 18}{3 \times 2 \times 1}$).
- Combinations with Repetition: Used when selecting a group of identical items or selecting from distinct items with replacement where order doesn't matter. For example, distributing 3 identical candies to 20 employees where each can get more than one (Stars and Bars method).
In our holiday ticket distribution problem, the tickets are distinct, and the recipient (employee) is chosen for each ticket, independently and with replacement (employees can be chosen multiple times). This aligns with the $r^n$ formula where $r$ is the number of choices for each item (employees) and $n$ is the number of items (tickets).
Important Questions from Permutations and Combinations
What is the number of 6-digit numbers that can be formed only by using 0, 1, 2, 3, 4 and 5 (each once); and divisible by 6 ?
Consider the following statements for a fixed natural number n:
1. C(n, r) is greatest if n = 2r
2. C(n, r) is greatest if n = 2r - 1 and n = 2r + 1
Which of the statements given above is/are correct ?
Let x be the number of permutations of the word ‘PERMUTATIONS’ and y be the number of permutations of the word ‘COMBINATIONS’. Which one of the following is correct ?
What is the number of ways in which 3 holiday travel tickets are to be given to 10 employees of an organization, if each employee is eligible for any one or more of the tickets?
A polygon has 44 diagonals then the number of its sides is
