Your instructor expects that you know about

permutations &

combinations.

If you study the two webpages I listed above then you will how those are related to counting problems.

$_n\mathcal{P}_k=\dfrac{n!}{(n-k)!}$ permutations are about

**order** in arrangements.

$_n\mathcal{C}_k=\dbinom{n}{k}=\dfrac{n!}{(k!)(n-k)!}$ combinations are about

**content** of arrangements.

There are twenty-six letters in the English alphabet. How many ten letter "words" have all letters different.

Well that is about order: $_{26}\mathcal{P}_{10}=\dfrac{26!}{(26-10)!}=19275223968000$

How many ten letter subsets are there? Well that's about content" $_{26}\mathcal{C}_{10}=\dbinom{26}{10}=\dfrac{26!} {(10!)(16)!}=5311735$

For the Newton's binom: $\displaystyle {(a + b)^n} = \sum\limits_{k = 0}^n {\left( {\begin{array}{*{20}{c}} n\\ k \end{array}} \right){a^{n - k}}{b^k}} $

Using above if $a=1~\&~b=1$ we get the really interesting result: $\displaystyle {2^n} = \sum\limits_{k = 0}^n {\left( {\begin{array}{*{20}{c}} n\\ k \end{array}} \right)} $ which happens to be the number of subsets on any set containing $n$ elements.

Thus there are a total of $2^{26}$ subsets of the English alphabet.