probability ,please correct me Finding Number of cases?
Suppose we have 40 cards numbered 1 through 40 and we want to pick 6 of them one at a time such that order of picking card “does matter”.
A. probability of numbers 1 and 2 are not among those cards?
B. probability of at least numbers one of numbers 10, 11, 12 are among those cards?
C. probability of at least numbers two of numbers 20, 21, 22 are among those cards?
Please correct me . Thanks
a)  P(38,6)/P(40,6)
b)  1- [p(37,6)/p(40,6)]
I have no idea for part c

A and B are fine.  Note that it isn't important whether the order matters or not.  If you say it doesn't matter, all your P's become C's, but the answer is the same.  For C, you need the inclusion-exclusion principle.  You can calculate the probability that 20 and 21 are both among the cards, then 21 and 22, then 20 and 22, and add them up.  The problem is you have triple counted the cases where all three are included.  So you need to subtract twice the chance that all three are included and you have your answer.
