Question 472787
The letters of DIRECTOR are arranged at random. What is the probability of vowels occupying ONLY even places.
Kindly provide the detailed explanation.
<pre>
DIRECTOR has 5 consonants (C) and 3 vowels (V).  There are 4 even places 
but only 3 vowels.  So we choose 3 of the 4 even places to put the 3
vowels.  So we can chose 3 of the 4 even places to  

So it has one of these C(4,3) forms:

CVCVCVCC, CVCVCCCV, CVCCCVCV, or CCCVCVCV

Every one of these C(4,3) ways to place the 3 vowels leaves the 
remaining 5 consonants D,R,C,T,R
to arrange around them.  The 2 R's are indistinguishable, so for each 
of the C(4,1) ways there are {{{5!/2!}}} ways to rearrange D,R,C,T,R
around the three vowels. So the numerator of the probabiliy is:

C(4,3)*{{{5!/2!}}} = 240

The denominator is the number of ways to arrange the letters DIRECTOR
in any manner, with the 2 R's indistinguishable:

{{{8!/2!}}} = 20160

So the desired probability is {{{240/20106}}} = 1/84


Edwin</pre>