Choosing numbers at random from a set $\{0,1,2, \cdots 9\}$ We choose numbers at random from a set $\{0,1,2, \cdots 9\}$ (without replacement), four numbers a,b,c,d. What is the probability that:
a)a is the greatest value of those we have chosen
b) a,b,c,d is an ascending sequence
My question is first and foremost about the logic i.e. i want to know how to calculate those probabilities but i am much more interested why this is the probability we are looking for.
I know that for certain:
$$|\Omega|=10\times9\times8\times7$$
 A: a) We are choosing numbers one at a time. Whatever collection of numbers we chose, by symmetry the probability the first chosen number was biggest is $\frac{1}{4}$.
We can also make a calculation based on your sample space $\Omega$. As you pointed out, this sample space has $(10)(9)(8)(7)$ outcomes. We now count the favourables.
There are $\binom{10}{4}$ ways to choose a collection of $4$ numbers. For every such choice, there are $3!$ ways to arrange the numbers so that the biggest is first, for the others can be arranged in $3!$ ways. Thus the number of favourables is $\binom{10}{4}3!$.
Finally, for the probability, divide the number of favourables by $|\Omega|$. After some simplification, we get $\frac{1}{4}$. The simplicity of the answer strongly hints that there must be a nicer way.
b) The same symmetry argument shows that the probability our numbers are in increasing order is $\frac{1}{4!}$. For whatever numbers we chose, all $4!$ orderings are equally likely.
We can also work with the sample space $\Omega$. In this case, there are $\binom{10}{4}$ favourables. 
