Probability of choosing 3 numbers from a set I am given a set $$A=(1,2...,8)$$ and I need to choose 3 numbers from set $A$ without replacement then what would be probability that minimum number is $3$ and maximum is $6$ ? Firstly I chose both $3,6$ and then one of the number from $(4,5)$ or first from $(4,5)$ and then $3,6$ or one at a time
$$P(E)= \frac{1}{^8C_2}.\frac{^2C_1}{^6C_1} + \frac{^2C_1}{^8C_1}.\frac{1}{^7C_2} + \frac{1}{8}.\frac{1}{7}.\frac{2}{6}$$ but I think my answer may not be correct , how to do this right?
 A: There are $\binom{8}{3}$ equally likely ways to choose $3$ numbers from the $8$. Of these choices, $2$ are "favourable," namely $3,4,6$ and $3,5,6$. So the probability is $\frac{2}{\binom{8}{3}}$.
A: Just Count:
The number of ways to choose $3$ out of $8$ values is $\dbinom83=\color{red}{56}$.
There are $\color{red}{2}$ ways to choose them with $3$ being the smallest and $6$ being the largest:


*

*$3,4,6$

*$3,5,6$


Hence the probability is $\color{red}{\dfrac{2}{56}}$.
A: Interpreting as "minimum number is 3" meaning, no number smaller than 3 appears, and "maximum number is 6" meaning, no number larger than 6 appears (allowing 456 and 345 as allowable answers as well)
You have the option of approaching this via a combinations argument and you also have the option of approaching this via a permutations argument.  In either scenario, we want to count the number of ways of choosing three numbers that satisfy the conditions requested and divide by the number of ways of choosing three numbers with no restriction.
Since there are four numbers available to choose from if we try to satisfy the condition that the smallest number is 3 and the largest number is 6:
$P(E) = \frac{|E|}{|S|}=\frac{\binom{4}{3}}{\binom{8}{3}} = \frac{4\cdot 3\cdot 2}{8\cdot 7\cdot 6} = \frac{1}{14}$
