How many permutations are there of the letters in word: Statistics?, with restriction. How many permutations are there of the letters in a word "statistics", such that the word starts with "s" and end with "s".
Is one of the following correct?
$$\frac{10!}{3! \cdot 3! \cdot 1! \cdot 2! \cdot 1!} = 50400$$
or
$$\frac{8!}{1! \cdot 3! \cdot 1! \cdot 2! \cdot 1!} = 3360$$
 A: Think of it like this: $\;s\;x_1\;x_2\;x_3\;x_4\;x_5\;x_6\;x_7\;x_8\;s\;$ with the $x_i$'s come from the set $\{S,t,a,t,i,t,i,c\}$. 
This is clearly a permutation of $8$ letters of whom $3$ t's and $2$ i's are redundants. So you have $\dfrac{8!}{2!\times 3!}$ words that starts and ends with $s$.
A: There's a technique in probabilistic reasoning that is very useful in questions like this where you have "...keep $x$ fixed..." or "...at least 1 in each container...".
The trick is to first "set up" the scenario by distributing the fixed restrictions and then counting the possibilities from what you have remaining to work with. For example applied to your question you need it to start and end with an S. So from our metaphorical bag of letters, let us take out two S and place them one on each end. Now all that we need to count are the permutations of the remaining 8 letters in our bag (I'll leave this up to you).
This generalises nicely to the problem of putting $r$ objects from your collection of $n$ into $r$ particular locations leaving you with the smaller problem of just permuting $n-r$ objects.

As an other unrelated example to get my message across about the technique (because ultimately, it's the technique that you need to take away from this) consider:
Ways to distribute $n$ objects into $k$ boxes such that no box is empty. Objects indistinguishable/distinguishable and each box can hold any number of objects.
Using the technique, put exactly one object in each box and then count the number of ways to distribute $n-k$ objects into $k$ boxes. For the case where the objects are distinguishable you need to also count the number of ways to choose and permute the $k$ objects you use to fill.
