Three dice having digits 1,2,3,4,5 and 6 on their faces are marked . I,II and III and rolled . Let x,y and z represent the number on die-I die -II and die -III respectively. What is the number of possible outcomes such that `x gt y gt z `?

Options

A. 14
B. 16
C. 18
D. 20

Answer

To solve the problem of finding the number of possible outcomes such that \( x > y > z \) when rolling three dice, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**: We have three dice, each with faces numbered from 1 to 6. We need to find the number of outcomes where the number on die I (x) is greater than the number on die II (y), which in turn is greater than the number on die III (z).

2. **Identifying Possible Values**: The possible values for \( x, y, z \) are from the set {1, 2, 3, 4, 5, 6}.

3. **Choosing Distinct Values**: Since \( x, y, z \) must be distinct and ordered such that \( x > y > z \), we can first choose any 3 distinct numbers from the set of 6 numbers.

4. **Calculating Combinations**: The number of ways to choose 3 distinct numbers from 6 is given by the combination formula \( \binom{n}{r} \), where \( n \) is the total number of items to choose from, and \( r \) is the number of items to choose. Here, \( n = 6 \) and \( r = 3 \):
\[ \binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \]

5. **Arranging the Chosen Values**: For each combination of 3 distinct numbers, there is exactly 1 way to arrange them such that \( x > y > z \). Therefore, we do not need to multiply by any additional factors for arrangements.

6. **Final Count of Outcomes**: Thus, the total number of outcomes where \( x > y > z \) is simply the number of ways to choose the 3 distinct numbers, which we calculated as 20.

### Conclusion:
The total number of possible outcomes such that \( x > y > z \) is **20**.