permutations without any successive digits

Question

I need some help with the following problem.

Consider the permutations of the set $ \{1, \dots , n \}$. What is the probability to find a permutation in which the digits $1$, $2$, $3$ are not successive. In other worlds what is the probability to occur a permutation of the form $$(\dots, a_1 , \dots, a_2, \dots, a_3, \dots) \quad, \{ 1,2,3 \} = \{ a_1, a_2, a_3 \} $$ ,while the dots before $a_1$ and after $a_3$ could be omitted. Obviously, every single permutation of the set has the same probability to occur.

The sample space, $\Omega$, of our random experiment is the set of all permutations of $ \{1, \dots , n \}$, so $$N_{\Omega} = n!$$

Now lets name $A$ the event described above. I should count how many permutations, of this specific form, there are. I will count them, in three steps, using the multiplication principle.

step 1

Consider the important digits, $1, 2, 3$, as ones and all the other as zeros. In step one, I will find in how many ways the ones could be arranged.

Let $y_1$ be the number of zeros before the $a_1$, with the same logic I define $y_2$ and $y_3$, finally let $y_4$ be the number of zeros after $a_3$.

It should be $$\sum y_i=n-3$$ with $$ y_1, y_4 \geq 0\, \text{ and }\,y_2, y_3 \geq 1 $$

Setting $$ x_1 = y_1 \quad x_2 = y_2 - 1 \quad x_2 = y_2 - 1 \quad x_4 = y_4 $$ I get the equivalent system $$\sum x_i = n-5\quad, x_i\geq0$$

The number of the integer solutions of the above system is actually the ways the ones could be arranged. In fact there are $$\binom{4 - 1 + n-5}{4-1}=\binom{n-2}{3}=\frac{(n-2)!}{(n-5)! \cdot 3!}$$ ways.

step 2

For every selection of the positions of the ones there are $3!$ ways to arrange the digits $1, 2, 3$

step 3

Finally all the other digits could fill the zeros positions in $(n-3)!$ ways

So, using the the multiplication principle I get $$N_{A} = \binom{n-2}{3}(n-3)! \cdot 3! = \frac{(n-2)!}{(n-5)! \cdot 3!} (n-3)! \cdot 3! $$

and $$P(A)=\frac{N_A}{N_\Omega} = \frac{(n-2)!\cdot(n-3)!\cdot 3!}{n! \cdot(n-5)! \cdot 3!} = \frac{(n-2)!}{n!}\frac{(n-3)!}{(n-5)!} = \frac{(n-4)(n-3)}{(n-1)n}$$

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Is it everything alright with my solution? Are their any misconceptions? Is there another way to tackle the problem?

Thank you for your time

Answer 1

$\underline{Another\;way\;to\;tackle\;the\;problem}$

Your answer is correct, but since probability has been asked for,

you could consider using $\dfrac{\binom{n-2}3}{\binom{n}3}$ to get the same answer.

[The numerator places the $3$ special $\#s$ in the gaps (including ends) of the remaining numbers ]

Answer 2

As you found, the focus set $\{1,2,3\}$ can be arranged $3!$ ways, the remainder $(n-3)!$ ways and the two part-sets can be interleaved by choosing $3$ of the $n{-}2$ gaps between and around the remainder elements in $\binom {n-2}3$ ways, giving the probability: $$ 3!(n-3)!\binom {n-2}3 \frac 1 {n!} = \frac {3!(n-3)!(n-2)!} {n!(n-5)!3!} = \frac{(n-2)!}{n!} \frac{(n-3)!}{(n-5)!} =\frac{(n-3)(n-4)}{n(n-1)}$$