Question 1097692
<pre>
First we calculate how many indistinguishable 
permutations of the 16 letters:

A,D,D,E,I,L,L,M,M,O,O,O,R,R,T,V

If the letters were distinguishable the 
answer would be 16!

However there are:
 
2 indistinguishable D's,
2 indistinguishable L's,
2 indistinguishable M's,
3 indistinguishable O's,
2 indistinguishable R's.

So we must divide the 16! by 2!2!2!3!2!

{{{16!/(2!2!2!3!2!)}}}

That works out to be 217945728000.

So the probability is 1 way out of 217945728000 or {{{1/217945728000}}}

Edwin</pre>