Question 252101
let the probability of face with 1 dot turning up be 1k
the probability of face with 2 dots turning up  2k
the probability of face with 3 dots turning up  3k

and so on

but since an outcome should be any one the face the sum of all these should be 1
so,1k+2k+3k+4k+5k+6k=1

that is 21k=1

we get k= 1/21

now the probability that each face turns up exactly once in 6 throws is 

1k*2k*3k*4k*5k*6k
that is {{{Factorial 6 * k^6}}}

so the answer is {{{720/(21)^6}}} or Factorial 6/21^6