In $22$ Randomly assigned letters, what is the probability that at least one of three words will appear? We write $22$ Different letters in a sequence, each must appear but only once.
The letters are $x_1,x_2,x_3,...,x_{22}$.
What is the probability that at least one of the following words will appear: $x_1x_2x_3, x_4x_5x_6, x_7x_8x_9$

What I did:
So it's the probability that we will get at least one of the three words in the sequence of $22$ letters, each word built from $3$ Different letters.
Now the probability that one of them will surely appear is:
Choose one of the three words - $\binom{3}{1}$ - put them in the sequence, we have left with $(22-3)$ Letters to order randomly, namely: $19!$ possibilities, now divide that by the total number of sequences of $22$ Different letters, each appear once = $22!$ possibilities
The rationale is that if we surely have $1$ Of the words in the sequence, its enough to gerenty that at least one will be there.
Therefore, I think the probability that at least one word will appear is:
$$
\frac{\binom{3}{1} \cdot 19!}{22!}.
$$
Surely, I am wrong.
I know that usually, questions like this are solved by looking at the complementary probability, namely, the $1 - P$(none of the words appear), yet I didn’t succeed in calculating this $P$ and moreover, why what I did is wrong?
Thanks.
 A: As lulu indicated in the comments, you have to apply the Inclusion-Exclusion Principle.  As G Cab indicated in the comments, the reason your attempt to use the Inclusion-Exclusion Principle did not work is that you have to take the position of the words into account.
Let $A$ denote the set of permutations which include the word $x_1x_2x_3$, let $B$ denote the set of permutations which include the word $x_4x_5x_6$, and let $C$ denote the set of permutations which include the word $x_7x_8x_9$.  Then the number of permutations which include at least one of these three words is
$$|A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |A \cap C| - |B \cap C| + |A \cap B \cap C|$$
$|A|$:  If a permutation includes the word $x_1x_2x_3$, then we have $20$ objects to permute, the word and the other $22 - 3 = 19$ letters.  Since the $20$ objects are distinct, they can be permuted in $20!$ ways.
By symmetry, $|A| = |B| = |C|$.
$|A \cap B|$:  If a permutation includes the words $x_1x_2x_3$ and $x_4x_5x_6$, we have $18$ objects to permute, the two words and the other $22 - 2 \cdot 3 = 16$ letters.  Since the $18$ objects are distinct, they can be permuted in $18!$ ways.
By symmetry, $|A \cap B| = |A \cap C| = |B \cap C|$.
$|A \cap B \cap C|$:  If a permutation includes the words $x_1x_2x_3$, $x_4x_5x_6$, and $x_7x_8x_9$, then we have $16$ objects to permute, the three words and the other $22 - 3 \cdot 3 = 13$ letters.  Since the $16$ objects are distinct, they can be permuted in $16!$ ways.
Hence, the number of favorable cases is
$$\binom{3}{1}20! - \binom{3}{2}18! + \binom{3}{3}16!$$
Dividing by the $22!$ possible permutations of the letters gives the desired probability.
