how many ways are there to order the word LYCANTHROPIES when C isn't next to A, A isn't next to N and N isn't next to T a. total number of ways to order 13 letters in a word: 13! 
b. Number of ways for CA/AC: 2*12! 
Number of ways for AN/NA: 2*12! 
Number of ways for NT/TN: 2*12! 
Total: 3*2*12! 
c. Number of ways for CAN/NAC: 2*11! 
Number of ways for ANT/TNA: 2*11! 
Total: 2*2*11! 
d. Number of ways for CANT/TNAC: 2*10! 
using the inclusion–exclusion principle I got:
13! -(3*2*12!) + (2*2*11!)-(2*10!). $$-$$
the textbook solution is: 13! -(3*2*12!) + (2*2*11!) +(2*2*11!)-(2*10!).
I dont know what I am missing. Any help is greatly appreciated. 
 A: You overlooked the case in which there are two disjoint pairs of prohibited adjacent letters.  
Strategy:  There are $13$ distinct letters in LYCANTHROPIES, so there are $13!$ arrangements of its letters.  From these, we must subtract those arrangements in which there are one or more prohibited pairs. 
A prohibited pair of adjacent letters:  You correctly calculated that there are $3 \cdot 12!2!$ such arrangements in part b of your work.
Two prohibited pairs of adjacent letters:  This can occur in two ways.  


*

*Two overlapping pairs of adjacent letters:  This means that you have three consecutive letters, namely CAN, NAC, ANT, TNA.  Assuming you meant NAC rather than NA$\color{red}{\text{T}}$, you correctly calculated that there are $2 \cdot 2 \cdot 11!$ such arrangements in part c of your work.

*Two disjoint pairs of adjacent letters:  This is the case you overlooked.  


The two disjoint pairs are CA/AC and NT/TN.  We have $13$ letters in total, so there are $11$ objects to arrange, the block containing A and C, the block containing N and T, and the other nine letters.  The objects can be arranged in $11!$ ways.  In each block, there are $2!$ ways to arrange the letters within the block.   Hence, there are $11!2!2!$ arrangements of this type.  This is the missing term.
Three pairs of prohibited adjacent letters:  This means that you have four consecutive letters, namely CANT, TNAC.  You correctly calculated that there are $2 \cdot 10!$ such arrangements in part d of your work.
Hence, by the Inclusion-Exclusion Principle, the number of admissible arrangements is 
$$13! - 3 \cdot 12!2! + 2 \cdot 2 \cdot 11! + 11!2!2! - 2 \cdot 10!$$
A: In part C of your calculation, you have forgot to include/exclude the option of: C next to A and N next to T
For this option you'll have $2 * 2 * 11!$ ways, which alongside with your final answer matches what the textbook states.
