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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 > y > z?

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 > y > z? Correct Answer 20

Calculation:

Let x, y and z represent the number on die-I, die-II and die-III respectively.

Given: x > y > z

It means the smaller number is shown on Die-III (z) than Die-II (y)

 

Possible cases

Die-III

(Z)

Die-II

(y)

Die-I

(x)

Total No. of Ways

 

I

 

1

2

3, 4, 5, 6

4

3

4, 5, 6

3

4

5, 6

2

5

6

1

 

II

 

2

3

4, 5, 6

3

4

5, 6

2

5

6

1

III

3

4

5, 6

2

5

6

1

IV

4

5

6

1

 

Total number of ways = 4 + 3 + 2 + 1 + 3 + 2 + 1 + 2 + 1 + 1 = 20

∴ Option 4 is correct.

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