"Statistics": How many of these begin and end with the letter "S"? 
How many distinct permutations are there of the letters
  in the word “statistics”? How many of these begin
  and end with the letter s?

The first part of the question I do understand. You have to use permutation with identical items. This is based on the number of letters.
$$\binom{10}{3,3,1,2,1} = 50400$$ 
Yet for the second part I am confused as to what the directions means. It says that how many letters begin and ends with letters  s so does one eliminate $2$ s and calculate this problem normally? 
BONUS:
 If so using a similar example how does one find out if how many of these begin and end with the letters m for the word
 "mathematicsman"?
 A: Oh my, such confusion! Let's try to simplify this, yet keep its essence.
Instead of statistics, let's use stats. There are 30 unique
arrangements, of which 3 have s at each end:
$$ \frac{5!}{2!\,2!} = 30\; \; \text{and} \; \; 3!/2! = 3.$$
So we can get the second number by removing s from each end
and dealing with what remains.
This means that in a random permutation. the probability of
getting an s at each end, should to 0.1.
In the simulation program below, I have avoided the messiness
of dealing with character strings in R, by substituting numbers
for letters in stats (1 represents s). A million random
permutations ought to give 2-place accuracy, so the answer
substantially matches the theoretical value. 
 stats = c(1,1,2,2,3)
 n = length(stats)
 m = 10^6; x = numeric(m)
 for (i in 1:m) {
  perm = sample(stats,n)
  x[i] = (perm[1]==1 & perm[n]==1) }
 mean(x)
 ## 0.099806

Now let's use mamam instead of mathematicsman. Again here, we can
use the standard method find $\frac{5!}{3!\cdot 2!} =10$ as the number of unrestricted, distinguishable permutations. If we ignore two of the (indistingusihable) m's, then we
have the 3 arrangements of ama. So the probability
a random permutation has m's at both ends should be $3/10 = 0.3.$ The simulation below
confirms this.
 mamam = c(1,1,1,2,2)
 n = length(mamam)
 m = 10^6; x = numeric(m)
 for (i in 1:m) {
   perm = sample(mamam,n)
   x[i] = (perm[1]==1 & perm[n]==1) }
 mean(x)
 ## 0.300217

A: I think it is (3 choose 1) times (9 choose 2,3,1,2,1). 
My reasoning is that you force an 's' at the front of your word in three different ways and then shuffle the rest of the letters around. 
A: Since 'statistics' must start and end with an s this reduces the total permutations of statistics (-tatistic-) to 8!. For distinct permutations - you need to account for the 3! ways you can rearrange the 3xt in -tatistic- and the 2xi (2!).
