How many arrangements of INCONSISTENT are there in which NE appear consecutively or NO appear consecutively, but not both NE and NO are consecutive?

Question

How many arrangements of INCONSISTENT are there in which NE appear consecutively or NO appear consecutively but not both NE and NO are consecutive?

Answer 1

Cases (NENO)

Arrangement of 9 elements NENO,I,N, etc... Note that if an element appears $k$ times, every specific arrangement will be counted k! times. We need to divide the number of arrangement by k!.

$\frac{9!}{2!2!2!}$

Cases (NONE or NENO)

Getting NONE has the same probability and correspond to disjoint cases because O and E only appears once in the original word.

$2 \times \frac{9!}{2!2!2!}$

Cases (NO)

Same logic as before with 11 elements

$\frac{11!}{2!2!2!2!}$

Cases (NO or NE)

NO and NE are overlapping you can have both. Therefore in order to simply double the combination we had before we need to remove the cases when we have both NO and NE.

$2 \times \frac{11!}{2!2!2!2!} - \frac{10!}{2!2!2!}$

Cases (NO and NE)

10 elements NO,NE,I,N,etc...

$\frac{10!}{2!2!2!}$

Cases where you have (NE or NO) and ((not NONE) and (not NENO))

$2 \times \frac{11!}{2!2!2!2!} - \frac{10!}{2!2!2!} - 2 \times \frac{9!}{2!2!2!}$

I believe this problem is simple enough to be addressed this way. For more complex problem refer to combinatorics polynomials like Laguerre and rook polynomials.