Question 548357
How many arrangements can be formed of the letters EXAMINATION so that vowels occupy odd places.
<pre>
Look at an arbitrary arrangement with vowels in the odd
positions, say, this one:

AMANENITIXO 

is composed of the two arrangements

A A E I I O  

   and

 M N N T X

The number of distinguishable arrangements of A A E I I O is {{{6!/(2!2!)}}}
We have to divide by the factorial of each number of indistinguishable
letters. We divided by 2! because of the 2 A's and again by 2! because of
the 2 I's.

For each of those, the number of distinguishable arrangements of M N N T X is
{{{5!/2!}}}  We have to divide by the factorial of the number of
indistinguishable letters N. So we divided by 2! because of the 2 N's.

So the total number is  

{{{6!/(2!2!)}}}×{{{5!/2!}}} = 10800

Edwin</pre>