Probability problem ( shuffling cards ) Suppose we shuffle a deck of 10 cards, each bearing a distinct number from 1 to 10, to mix the cards thoroughly. We then remove three cards, one at a time, from the deck. What is the probability that we select the three cards in sorted (increasing) order?
 A: Let's first consider that we draw 3 cards without ordering them, then we order our 3 drawn cards. This is an equivalent problem and regardless of the cards we drew, there are 6 equiprobable orderings, only one of which yields an increasing order.
The probability is therefore 1/6.
A: Let E = event that we draw 3-cards that are in increasing order
Then P(E) = |E|/|S| 
Note we can use P(E)= |E|/|S| because each outcome is of equal probability.
First, we find the size of the sample space |S| as the total permutations of 3 cards out of 10 cards:
n!/(n-k)! = 10!/7! = 720. That's the easy part.
Second, what are the total # of 3-card orderings that are in increasing order? i.e. |E|
This is simply the total combinations of 3 cards out of 10 cards:
n!/[n!(n-k)!] = 10!/(3!7!) = 120
Why can we use the combinations formula? 
Well, when counting combinations, order doesn't matter, 
so {1,2,3} = {2,1,3} = {2,3,1} = {3,2,1} = {3,1,2} 
In any set of the same numbers, there is only one ordering where the numbers are increasing. 
Therefore, by counting the combinations, its the same as counting the only sorted order and disregard the non-sorted orders.  
P(E) = |E|/|S| = 120/720 = 1/6 
Please feel free to let me know if I made some logical shortcuts in the second part.
A: I used the approach of counting. Since we are drawing one card at a time: 
$S = 10 * 9 * 8$ (10 ways to pick first card, then 9, then 8) and 
$$E = \sum_{i = 1}^8 \sum_{j = i+1}^9 \sum_{k = j+1}^ {10}i.j.k$$
As somebody already pointed out, a combinatorial nightmare. Wondering how this shall turn out to be $10C3$
