Question

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? 

a. 14 

b. 16 

c. 18 

d. 20

Answer 1

Correct option d. 20

Explanation:

No. of possible out comes

= 10 + 6 + 3 + 1 = 20