Probability for range 
To determine the probability I guess I have to find out the range from the number 1 to 10. 
What will be the range here? Will it be 7?
How do I proceed from getting a range of 7 to finding the probability of getting range 7?
 A: Firstly, just to start you off, if I pick 4 discs at random and get say $$\{2, 5, 6, 9\}$$ here the range is $7 = 9 - 2$, between the smallest and largest. Ask yourself at least fleetingly, a detailed answer may not be the best way to proceed, how many ways can four discs have $7$ as their range?
A less sledge hammered clue, just in case, is to think about how many ways there are for two discs to be at each end of a range of 7.
My next clue would probably give away the answer.
My calculation gives the suggested answer of $3/14$, so I think I'm on the right track. If these clues don't kick you off, just ask and I'll elaborate.
Okay, I have to sign off soon, so I'll give a complete answer now just in case I don't get back in time.
Elaboration: $(1,8), (2,9), (3,10)$ are the only pairs with range $7$. To be the lowest and highest, the other two discs have to be in between. In each case there are six numbers in between. So now, how to count everything?
(1) There are ${{10}\choose{4}}$ ways of choosing 4 distinct discs.
(2) There are 3 ways to choose a range 7 pair.
(3) Given you have a range 7 pair there are ${{6}\choose{2}}$ ways to choose the two which must have a value in between the range 7 pair
Thus
\begin{align*}
P(R7) &= \frac{3{{6}\choose{2}}}{{{10}\choose{4}}}\\
&= \frac{3 (6!\,/\,(4! 2!))}{10!\,/\,(6! 4!)} \\
&= \frac{3 (6! 6! 4!)}{10! 4! 2!} \\
&= \frac{3}{14}
\end{align*}
