Find $P\{\min_{i \neq j}|R_i-R_j| \geq d\}$, where $R_1,\ldots,R_n$ are uniform on line with length $L$ If n points $R_1,\ldots,R_n$ are picked independently and with uniform density on a straight line of length L, find the probability that no two points will be less than distance d apart; that is, find
$$P\{\min_{i\neq j}|R_i-R_j| \geq d\}$$
Attempt: 
The hint given in is to find $P\{\min_{i \neq j}| R_i-R_j| \geq d\}$, $R_1<R_2<\cdots<R_n$}, and to show that the region of integration defined by this event is 
$$x_{n-1}+d \leq x_n \leq L$$
$$x_{n-2}+d \leq x_{n-1} \leq L - d$$
$$x_{n-3}+d \leq x_{n-2} \leq L - 2d$$
$$\vdots$$
$$x_{1}+d \leq x_{2} \leq L - (n-2)d$$
$$0 \leq x_{1} \leq L - (n-1)d$$
I'm also wondering if it's possible to incorporate Chebychev's inequality:
$P(g(X) \geq r) \leq \dfrac{Eg(X)}{r}$, where r would be equal to d. 
Solution is given as $\left[1-\frac{(n-1)d}{L}\right]^n$, if $(n-1)d \leq L$, and 0 if $(n-1)d > L$.
 A: The random variable $(R_1,R_2,\ldots,R_n)$ is uniformly distributed
in $[0,L]^n$, so what you need to is compute the volume of
$$
A=\bigg\lbrace (x_1,x_2,\ldots,x_n)\in [0,L]^n \ \bigg| \ |x_i-x_j|\geq d
\ (i\neq j) \bigg\rbrace \tag{1}
$$
Now $A$ can be partitioned as $\bigcup_{\sigma\in{\mathfrak S}_n}A_{\sigma}$ where
$$
A_{\sigma}=\bigg\lbrace (x_1,x_2,\ldots,x_n)\in [0,L]^n \ \bigg| \ |x_i-x_j|\geq d
\ (i\neq j), \ x_{\sigma(1)}<x_{\sigma(2)}<\ldots<x_{\sigma(n)} \bigg\rbrace 
\tag{2}
$$
Now, on $A_{\sigma}$ let us change variables by putting
$$
y_{\sigma(k)}=x_{\sigma(k)}-(k-1)d \ (1\leq k \leq n) \tag{3}
$$
This linear transformation has determinant $1$, so the volume of $A_{\sigma}$ is the
same as the volume of $B_{\sigma}$ where 
$$
B_{\sigma}=\bigg\lbrace (y_1,y_2,\ldots,y_n)\in [0,L-(n-1)d]^n \ \bigg| \  \ y_{\sigma(1)}<y_{\sigma(2)}<\ldots<y_{\sigma(n)} \bigg\rbrace 
\tag{2}
$$
It follows that the volume of $A$ is the same as the volume of
$B=\bigcup_{\sigma\in{\mathfrak S}_n}B_{\sigma}$. Now $B=[0,L-(n-1)d]^n$,
so its volume is $(L-(n-1)d)^n$ and we are done.
