Personnel selection. Suppose that 6 female and 5 male applicants have been successfully screened for 5 positions. If the 5 positions are filled at random from the 11 finalists, what is the probability of selecting (A) 3 females and 2​ males? ​(B) 4 females and 1​ male? ​(C) 5​ females?

Data analysis 
Note : As per authoring guidelines, when multiple subparts are posted, then only first three subparts are to be answered. 
Given there is selection for a Personnel. 
And there are 11 finalists in which 6 are female and 5 are female. 
And there are only 5 posts vacant. 
To calculate the probability of selecting an item as described below.
The total number of ways:
Total number of ways of selecting 5 people from 11 finalists is given by, 
$11P5$ 
$=11\times 10\times 9\times 8\times 7$ 
$=55.440$ 
Probability is given by the ratio of number of favourable ways to the total number of ways. 
Note : 
$nPr=n\frac{!}{n-r}!$ 
1) 3 females and 2​ males 
3 females selected from 6 female finalists in $6P3$ ways 
$=6\times 5\times 4$ 
$=120$ 
2 males selected from 5 male finalists in $5P2$ ways 
$=5\times 4$ 
$=20$ 
Favourable ways $=120\times 20=2400$ 
Probability $=2400/55.440$ 
$=0.0432900433$ 
The probability of selecting 3 females and 2 males is 0.04329 
(Rounded to 5 decimals) 
2) 4 females and 1 males 
4 females selected from 6 female finalists in $6P4$ ways 
$=6\times 5\times 4\times 3$ 
$=360$ 
1 male selected from 5 male finalists in $5P1$ ways 
$=5$ 
Favourable ways $=360\times 5=18000$ 
Probability $=1800/55.440$ 
$=0.0324675325$ 
The probability of selecting 4 females and 1 male is 0.0324675 
(Rounded to 7 decimals) 
3) 5 females 
5 females selected from 6 female finalists in $6P5$ ways 
$=6\times 5\times 4\times 3\times 2$ 
$=720$ 
Favourable ways = 720 
Probability $=720/55.440$ 
$=0.012987013$ 
The probability of selecting 5 females is 0.012987 
(Rounded to 6 decimals)