combinatoricspermutations
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.
The six distinct letters of the word SUNDAY can be arranged in $6!$ ways along a line. However, if they are arranged along a circle, the six permutations SUNDAY, UNDAYS, NDAYSU, DAYSUN, AYSUND, YSUNDA obtained by successive $60^\circ$ rotations are indistinguishable. 
More generally, the six permutations that can be obtained by rotations of $0^\circ$, $60^\circ$, $120^\circ$, $180^\circ$, $240^\circ$, and $300^\circ$ are indistinguishable. Hence, there are
$$\frac{6!}{6} = 5!$$
indistinguishable clockwise permutations of the word SUNDAY. By the same reasoning, there are $5!$ anti-clockwise permutations of the word SUNDAY, each of which corresponds to reading a given clockwise permutation in reverse order. Since we are distinguishing between clockwise and anti-clockwise arrangements, we count a given permutation and its reflection separately. Therefore, there are $5!$ circular permutations of the word SUNDAY.
As you observed, there are $$\frac{6!}{3!} = 120$$ distinct arrangements of the six letters of the word ALASKA in a line. Since there are six indistinguishable arrangements produced by rotations around the circle, there are $$\frac{6!}{6 \cdot 3!} = 20$$ distinct clockwise arrangements of the letters of the word ALASKA. Since we are treating reflections as distinguishable, there are $20$ distinguishable circular permutations of the word ALASKA.
Yes, these are all correct - standard approaches to solving these types of problems.
Best Answer
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
$$|\bar{A_1}\cap\bar{A_2}\cap\bar{A_3}| = \frac{11!}{2!2!2!} - 3\cdot\frac{9!}{2!} + 3\cdot\frac{7!}{2!}$$