# We arrange 12-letter words having at our disposal five letters a, four letters b and three letters c

We arrange 12-letter words having at our disposal five letters a, four letters b and three letters c. How many words are there without any block $5×a$, $4×b$ and $3×c$. I need to use inclusion - exclusion principle. I counted all possible 12 letter words - $\frac{12!}{5!4!3!}$. Then words with block of letters:
only a blocks - $8\cdot \frac{7!}{4!3!}$
only b blocks - $9\cdot \frac{8!}{5!3!}$
only c blocks - $10\cdot \frac{9!}{5!4!}$.
Now I have to count all words, where block: a, b or a, c, or b, c appears together, but I don't now how.
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Perman7z
Step 1
If we let N denote the total number of distinguishable arrangements of aaaaabbbbccc, A denote the set of permutations with five consecutive as, B denote the set of permutations with four consecutive bs, and C denote the set of permutations with three consecutive cs, then, by the Inclusion-Exclusion Principle, the number of arrangements with no block of five consecutive as or four consecutive bs or three consecutive cs is
$\begin{array}{rl}|{A}^{\prime }\cap {B}^{\prime }\cap {C}^{\prime }|& =N-|A\cup B\cup C|\\ & =N-\left(|A|+|B|+|C|-|A\cap B|-|A\cap C|-|B\cap C|+|A\cap B\cap C|\right)\\ & =N-|A|-|B|-|C|+|A\cap B|+|A\cap C|+|B\cap C|-|A\cap B\cap C|\end{array}$
Step 2
You have correctly calculated N, |A|, |B|, and |C|.
$|A\cap B|$ We have five objects to arrange: aaaaa, bbbb, c, c, c. There are $\left(\genfrac{}{}{0}{}{5}{3}\right)$ ways to select three of the five positions for the three cs and 2! ways to arrange the remaining two distinct objects in the remaining two positions. Hence, there are
$\left(\genfrac{}{}{0}{}{5}{3}\right)2!=\frac{5!}{3!}$ such arrangements.
Alternatively, arrange the three cs in a line. This creates four spaces in which we can place the block aaaaa, two between successive cs and two at the ends of the row.
$◻c◻c◻c◻$
Once we have placed the block aaaa, we have four objects in a row, which creates five spaces in which to place the block bbb, three between successive objects and two at the ends of the row. Hence,
$|A\cap B|=4\cdot 5$
Can you finish the argument by counting $|A\cap C|$, $|B\cap C|$, and $|A\cap B\cap C|$?