Question 1203274
<pre>
a) There are 12 letters, and if these were all unique, there would be 12! unique ways to arrange them.   However, we must divide out the duplicate (non-distinct) arrangements.   For this, we note there are 2 A's, 2 R's, 2 N's, and 2 E's, each of which contributes 2! non-distinct arrangements:

    Number of unique arrangements =  {{{ 12!/ (2!*2!*2!*2!) }}} = 29937600


Part (b) can be done similarly.  Since the arrangements begin with EE, that effectively leaves 10 letters to arrange, and you'll need to remove duplicates in a similar way as in part (a).  It is as if the E's are removed and you repeat part (a) without the E's. 

Hint:  the answer will be  10! / (something)