What is the probability that Raj and Rana have atleast 3 persons between them,given raj and rana are standing in a row and there are 9 peoples. Question:

Raj and Rana are standing in a row. There are 9 persons including Raj and Rana. What is the probability that at least 3 people will stand between Raj and Rana?

My solution:
$9$ people can be arranged in $9!$ ways. Excluding Raj and Rana, out of $7$ people.


*

*Case 1: I select 3 people $^7C_3$ who can be arranged in $3!$ ways.
Remaining $(9-(3+2))=4$ people can be arranged in $4!$ ways
Therefore required arrangement=$^7C_3*3!*4!*2!$   ----$(i)$

*Case 2: I select 4 people $^7C_4$ who can be arranged in $4!$ ways
Remaining $(9-(4+2))=3$ people can be arranged in $3!$ ways
Therefore required arrangement=$^7C_4*4!*3!*2!$   ----$(ii)$

*Case 3: I select 5 people $^7C_5$ who can be arranged in $5!$ ways
Remaining $(9-(5+2))=2$ people can be arranged in $2!$ ways
Therefore required arrangement=$^7C_3*5!*2!*2!$   ----$(iii)$

*Case 4: I select 6 people $^7C_6$ who can be arranged in $6!$ ways
Remaining $(9-(6+2))=1$ people can -be arranged in $1!$ way.
Therefore required arrangement=$^7C_6*6!*1!*2!$   ----$(iv)$

*Case 5: I select 7 people $^7C_7$  which can be arranged in $7!$ ways
Remaining $(9-(7+2))=0$ people can be arranged in $1!$ ways
Therefore required arrangement=$^7C_7*7!*1!*2!$   ----$(v)$
Adding equations $\frac{(i)+(ii)+(iii)+(iv)+(v)}{9!}$ we get the  required probability.
Is it a correct approach and if there exists any shorter method, could you please tell me?
Correct answer is $\frac5{12}$.
 A: Probably not the simplest solution, but I would solve it in the following way:
Let $X$ = Ray and $Y$ = Rana. There are of course $9!$ total possibilities to arrange $9$ people in a row.
Then when $X$ and $Y$ have their places set up, there are $7!$ possibilities to arrange all the other $7$ people. You have $15 \cdot 2$ possibilities to arrange $X$ and $Y$, so the requirments of the task are met, because when $X$ is arranged in a following way:
$X$ _ _ _ _ _ _ _ _ 
you have 5 possibilities to arrange $Y$ to the right of $X$. Each time you move $X$ to the right, the number of possibilities of $Y$ gets lower by $1$. So total number of possibilities for $Y$ when $X$ is moved to the right is $5+4+3+2+1 = 15$, and you can switch $X$ and $Y$, so the total number is $15 \cdot 2$. This yields the probability:
$$
\frac{30 \cdot 7!}{9!} = \frac{5}{12} \text{.}
$$
A: With stars and bars you can find the number of solutions of: $$a+b+c=4$$
where $a,b,c$ are nonnegative integers.
The answer to that is $\binom{6}{2}$.
Identify $a$ with the number of persons on left of both , $b+3$ with
the number of persons in between and $c$ with the number of persons
on the right of both.
In total there are $\binom92$ possibilities.
This gives the probability:$$\frac{\binom{6}{2}}{\binom{9}{2}}=\frac{5}{12}$$
A: As I have been asked to explain how I got the previous answer, this is the long explanation.
We have 9 people waiting in a queue. 
All the possible ways they could be arranged is 9!.
We assume it is equally probable for Raj and Rana to be at any position in the queue, so to compute the probability there are at least three people between them we need to know the number of pairs of positions that have at least three people between them.
Then since they are equally probable we divide this number by the number of all permutations.
Now let us compute. There are exactly $3$ people between two numbers $1\leq x < y \leq 9 \ \ x,y\in \mathbb{N} \ \ \iff y=x+4$
So $9-4=5$. Therefore $5$ is the greatest number with $3$ people between it and Ranna or Raj who we assume is $4$ positions ahead. Here we have $5$ favourable pairs then. which give us $10$ positions when we account for that it Raj and Rana can exchange places.
If there are $4$ people between Raj and Ranna we have by the same logic $4$ favourable pairs and $8$ positions.
Long story short we have $(1+2+3+4+5)2$ cases where there are at least 3 people between Raj and Rana.
Since we didn't care about the order in which the other people came we're not going to care about them when calculating all the positions Raj and Rana can possibly get in that line in.
One of them can be in $9$ positions, the other in the remaining $8$, so we have $72$ positions. 
Dividing we get $\frac{30}{72}=5/12$.
A: First look at the number of ways they can stand with no person in between them. There are exactly $2\cdot8\cdot7!$ such ways, since they always have to stand together and rest $7$ can stand anyway. Similarly, there are exactly $2\cdot7\cdot7!$ ways for Raj and Rana to stand such that there is exactly one person in between them, and $2\cdot6\cdot7!$ for two persons between them. So, the total is $42\cdot7!$ ways. So, the probability that there are at least $3$ people is $$1-\Pr(\text{two or less}).$$ Since there are $9!$ ways to arrange the $9$ persons, the answer is $$1-\frac{42\cdot7!}{9!} = \frac{5}{12}.$$ 
A: Treat Raj and Rana as white balls and the rest as black balls (hoping that they don't mind !)
Remove 3 black balls, then the white balls can be placed in ${6\choose 2}=15$ positions.
Insert the 3 removed black balls immediately to the right of the first white one.
Then only number the positions 1-9. This takes care of all favorable positioning of white balls.
Their unrestricted positioning is ${9\choose 2}= 36$
Thus Pr = $\frac{15}{36} = \frac{5}{12}$  
A: Somewhat less complicated answer. R&R can occupy 2 of the 9 places in 9C2 or 36 ways.
Atleast 3 people stand between R&R if the occupy following spaces


*

*15, 26, 37, 48, 59,  

*16, 27, 38, 49, 

*17, 28, 39, 

*18, 29, 

*19


Total 15 ways out of 36 ways or 5/12
