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

Question

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.

Answer 1

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.

1. 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.
2. 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 $$13! - 3 \cdot 12!2! + 2 \cdot 2 \cdot 11! + 11!2!2! - 2 \cdot 10!$$

Answer 2

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.