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 ?

Options

A. (a)60
B. (b)120
C. (c)500
D. (d)1000

Answer

To solve the problem of distributing 3 holiday travel tickets among 10 employees, where each employee can receive one or more tickets, we can follow these steps:

### Step 1: Understand the Problem
We have 3 identical tickets and 10 employees. Each employee can receive any number of tickets, including none. We need to find out how many different ways we can distribute these tickets.

### Step 2: Determine the Distribution Method
Since the tickets are identical and can be given to any of the 10 employees, we can think of this as a problem of distributing indistinguishable objects (tickets) into distinguishable boxes (employees).

### Step 3: Use the Stars and Bars Theorem
The Stars and Bars theorem is a common combinatorial method used to solve problems of distributing indistinguishable objects into distinguishable boxes. According to this theorem, the number of ways to distribute \( n \) indistinguishable objects into \( k \) distinguishable boxes is given by the formula:

\[ \text{Number of ways} = \binom{n+k-1}{k-1} \]

Where:
- \( n \) is the number of indistinguishable objects (tickets),
- \( k \) is the number of distinguishable boxes (employees).

### Step 4: Apply the Formula
In our case:
- \( n = 3 \) (the tickets),
- \( k = 10 \) (the employees).

Plugging these values into the formula gives us:

\[ \text{Number of ways} = \binom{3 + 10 - 1}{10 - 1} = \binom{12}{9} \]

### Step 5: Calculate the Binomial Coefficient
Now, we need to calculate \( \binom{12}{9} \):

\[ \binom{12}{9} = \binom{12}{3} = \frac{12!}{3!(12-3)!} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = \frac{1320}{6} = 220 \]

### Step 6: Conclusion
Thus, the total number of ways to distribute the 3 holiday travel tickets among 10 employees is **220**.