Question

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

Answer 1

Total cases (Sample Space) = $6*6*6*6*6*6 = 6^6$ ( each dice has six possiblities )

Of all these $6^6$ cases we are interested in only those where  each number $[1,2,3,4,5,6]$ appears exactly once.

In first dice, there is no restriction on number that we should not get, So we are good to go if any of $6$ numbers comes up. Now while rolling the second dice we are not interseted in the number we already got.( remember every number has to come exactly once). So, we are good to go if any of remaining $5$ numbers come up. Similarly for third dice $4$ cases and following the same logic $3,2,1$ cases for fourth, fifth and sixth dice respectively. 

So, total favourable cases(distinct number on each dice) = $6*5*4*3*2*1 = 6!$

P = $\frac{6!}{6^6} = 0.015432$

Answer 2

Answer should be: 6!/(6)^6

 

Favourable outcomes: 6!

Total outcomes : 6^6

Comments

Some favourable cases : $[1,2,3,4,5,6]$   $[1,3,2,4,5,6]$   $[1,4,3,2,5,6]$   $[1,5,6,2,3,4]$   $[1,6,5,2,3,4]$ $[1,3,6,5,2,4]$  $[1,5,6,4,3,2]$

Favourable cases are more than 6.

Modify your answer.