a)Three identical letters are not next to each other.
So I came up with a solution but I have no possibility to check whether it's correct so I've decided to post it here.
So |X| - number of all possible sequences,
|X| = $\frac{9!}{3!\cdot 3!\cdot 3!}$
|A|, |B| and |C| are those sequences in which either 3 A's, B's or C's are next to each other
|A| = |B| = |C| = $7\cdot \binom{6}{3}\cdot \binom{3}{3}= 140$
$|A\cap B|, |A\cap C| ,|B\cap C|$ are for use of Inclusion-exclusion
$|A\cap B|=|A\cap C|=|B\cap C|=7\cdot 4\cdot \binom{3}{3}= 28$
$|A\cap B\cap C|=1$
$|A\cup B\cup C|=|A|+|B|+|C|-|A\cap B|-|A\cap C|-|B\cap C|+|A\cap B\cap C|$
And the number we're looking for $=|X|-|A\cup B\cup C|$
Is that by any chance a correct solution?
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johny.bravo
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I think you have mistake in $A \cup B$ it is actually 20 – Jan 10 '18 at 21:40
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@lulu looked for it, couldn't find it, thank you! – johny.bravo Jan 10 '18 at 21:47
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Only mistake is in $\mid {A \cup B}\mid$ it is $5!\over {3!}$
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Did you mean $|A \cap B|$? and if yes, then where does this result come from? – johny.bravo Jan 10 '18 at 21:58
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The method is good. Two mistakes I can see:
In $A \cap B$, I'm not sure where your $7$ comes from. I'd do it this way: we have $5$ things to arrange: the $A$ block, the $B$ block, and three $C$s. Choose one slot for the $A$ block ($5$ choices) and one for the $B$ block ($4$ choices), and the rest are $C$s. So it should be $20$.
In $A \cap B \cap C$, you have to order the $3$ blocks, so it should be $3!$, not $1$
BallBoy
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