If six fair dice are rolled , what is the probability that each of the six numbers will appears exactly once ?

• chances of different digit appear on each dice=6!
  [suppose consider a case different digits appeared on all 6 dice, then remaining chances comes by arranging these 6 numbers in all possible ways without repeating.
  that is 6!]
  all possible chances of a number appear on each die=6^6
  so the probability required=6!/(6^6)=5!/(6^5)=(5*4*(3*2))/(6*6*6*6*6)=5/(9*6*6)
  =5/324...
• 11 years agoHelpfull: Yes(14) No(3)
• 
  
  
  Imagine you throw one after the other. You consider a throw as a success if the number is different from all previous numbers. You start with one. This is always a succes so P(first)=1=66. Your second throw is a success if one of the remaining 5 numbers shows, so P(second)=56. And so on. Since all the throws are independent, the total probability is the product of all separate probabilities:
  
  
  P(all numbers are different)=6/6⋅5/6⋅4/6⋅3/6⋅2/6⋅1/6
  
• 11 years agoHelpfull: Yes(4) No(1)