Permutations and Combinations of distinct discs and blanks Question: 15CDs, 12 have DATA, 3 have blanks. What is the probability that the 13th checked disk is the 10th disk with data given that the disks are distinct. 
So Total is 15!/3! because there are 15 disks that can be ordered, but since the 12 data disks are distinct, we only need to account for the 3 blanks, hence dividing by 3!
So how do we find the 13th disk to be the 10th disk with data
that means 3 blanks have to be before the 10th disks?
Is it just 13!/3!10!
 A: Method 1:  For the $13$th checked disk to be the $10$th disk with data, nine of the twelve disks with data and all three blank disks must be selected among the first $12$ disks selected, then a disk with data must be selected with the $13$th selection.  Since only disks with data are left if three blank disks are selected among the first $12$ disks, the probability that the $13$th disk will have data is $1$ if three blank disks have already been selected.  There are
$$\binom{12}{9}\binom{3}{3}$$
ways to select nine disks with data and three blank disks.  There are 
$$\binom{15}{12}$$
ways to select twelve of the fifteen disks.  Hence, the probability that the $13$th disk selected will be the $10$th disk with data is 
$$\frac{\dbinom{12}{9}\dbinom{3}{3}}{\dbinom{15}{12}}$$
Method 2:  If we arrange all $15$ disks in order, there are 
$$\binom{15}{3}$$
ways to choose the positions of the blank disks.  For the $13$th disk to be the $10$th disk with data, all three of the blank disks must occur in the first $12$ positions, which can occur in 
$$\binom{12}{3}$$
ways.  Hence, the desired probability is 
$$\frac{\dbinom{12}{3}}{\dbinom{15}{3}}$$
