I will give you an easier example only as a tool on how to calculate your question.

If you have 4 cards a,b,c,d and you take out 2 cards and you want it to be a,b

n=4 , k=2

Combinatipons of all the cards:

--------------------------------

You write all the combinations 4! = 24. [a,b,c,d] , [a,b,d,c] , [a,c,b,d],[a,c,d,b], [a,d,b,c],[a,d,c,b] ...[d,a,b,c].....[d,c,b,a]

now you need to count how many combinations contains a,b in the first 2 places. [a,b,c,d], [a,b,d,c] [b,a,c,d], [b,a,d,c]

combinations of a,b 2! [a,b] and [b,a]

combinations of c,d (4-2)! [c,d] and [d,c]

total combination of a,b in the first 2 placedes: 2!*2!=4

the probability of getting the first 2 cards as [a,b] or [b,a] (no mean for the order):

(number of combinations that meet your criteria) divided by (number of combinations)

4/24 = 1/6

Another option, using all the combinations to take two cards with no meaning to the order:

-------------------------------------------------------

number of combinations: 6 [a,b] , [a,c], [a,d], [b,c], [b,d], [c,d] n!/(k!*(n-k)!=4!/(2!*(4-2)!)=6 {n=4, k=2}

combination that meet your criteria: 1 [a,b]

the probability: 1/6