Arrangements of the word COMBINATION not containing CAN, BIN, NIB How many ways can we arrange the letters of the word COMBINATION, containing none of the patterns CAN, BIN and NIB?
Without restrictions, arranging the 11 letters with 2 O's, 2 I's and 2 N's is $\frac{11!}{2!2!2!} $ ways.
Where I'm stuck at now is how to calculate the arrangements with CAN, NIB and BIN, since there are 2 I's and 2 N's, and also each of the patterns contain letters in the other pattern.

Can anyone could help me out with this problem?
Thanks in advance.
 A: It's an application of the inclusion exclusion principle. We use the above to write properties $P_1, P_2$ and $P_3$ respectively as
$P_1$ - Pattern CAN occurs
$P_2$ - Pattern BIN occurs
$P_3$ - Pattern NIB occurs
Let $A_1, A_2, A_3$ be the sets conforming to each property. Now using the above stated principle we know that
$$|\bar{A_1}\cap\bar{A_2}\cap\bar{A_3}| = |S| - \sum_{i=1}^3|A_i| +\sum_{i,j}|A_i \cap A_j| - |A_1\cap A_2\cap A_3|$$
$A_1$ - Treat CAN as one entity and calculate the permutations of the entities (CAN),O,M,B,I,T,I,O,N which is equal to 
$$|A_1| = \frac{9!}{2!}$$
$A_2$ - Treat BIN as one entity and calculate the permutations of the entities C,O,M,(BIN),A,T,I,O,N which is equal to 
$$|A_2| = \frac{9!}{2!}$$
$A_3$ - Treat NIB as one entity and calculate the permutations of the entities C,O,M,(NIB),A,T,I,O,N which is equal to 
$$|A_3| = \frac{9!}{2!}$$
$A_1\cap A_2$ - Putting CAN and BIN together and calculating the permutations of the entities (CAN),(BIN),O,M,I,O,T which is equal 
$$|A_1\cap A_2| = \frac{7!}{2!}$$
$A_2\cap A_3$ - We cannot have BIN and NIB together as there's only one B in COMBINATION
$$|A_1\cap A_3| = 0$$
So, also $|A_1\cap A_2\cap A_3| = 0$ 
EDIT - As was suggested by the commenter, there's one extra case that I forgot to account for, which is CANIB. This will be a part of $A_1\cap A_3$ and is calculated by the number of permutations of (CANIB),O,O,T,I,N,M
$$|A_1\cap A_3| = \frac{7!}{2!} + \frac{7!}{2!}$$
Hence the total number of combinations become
$$|\bar{A_1}\cap\bar{A_2}\cap\bar{A_3}| = \frac{11!}{2!2!2!} - 3\cdot\frac{9!}{2!} + 3\cdot\frac{7!}{2!}$$
