Verify my solutions to counting problems I'm pretty sure I know the answers to these problems, but still want to double check.


*

*How many different ways are there of arranging all the letters of the string CALCULUSBOOK?
Solution: $12!$ since we want to arrange all the letters. other wise it would be $\dfrac{12!}{2!2!2!2!}$.

*What is the coefficient of $x^5$ in the expansion of $(3x - 1)^{11}$?
Solution: according to the binomial theorem the binomial coefficient would be $\dbinom{11}5$ since $(1-x)^k = \ldots+\dbinom{k}kx^k$
Please verify, correct or incorrect? 
 A: *

*You’ve misunderstood the import of the word all: it simply means that you’re to use all $12$ letters, not that you’re to consider the two $U$’s, for instance, to be distinguishable letters. Thus, the correct answer is in fact $$\dfrac{12!}{2!2!2!2!}\;.$$

*You forgot to take into account the coefficient of $3$. There will be $\binom{11}5$ terms of the form $(3x)^5\cdot1^6$, but in each of them the coefficient of $x$ is $3^5$, not $1$, so their sum has a coefficient of $\binom{11}53^5$, not $\binom{11}5$.
A: *

*No. This is the same number of ways to write CCALLUUSBOOK, which is equal to:


$$ \frac{12!}{2!^4} = \frac{12!}{16} = 29937600 $$
(Because there are four letters who appear twice, which are indistinguishable, and tweleve letters total)


*

*No. Use the more general binomial theorem:


$$(a+b)^n = \sum_{k=0}^n {n \choose k} a^k b^{n-k} $$
With $a=3x$, $b=-1$, and then evaluate the sum at $k=5$ (to get $x^5$).
A: When I approach these types of problems, I find it very helpful to reorder the string of letters as such:
$$
\text{CC}\\
\text{A}\\
\text{LL}\\
\text{UU}\\
\text{S}\\
\text{B}\\
\text{OO}\\
\text{K}.
$$
It may seem strange, but this may help prevent you from over counting. You have $12$ letters in all, and you have $4$ types of letters that have duplicates. Therefore you have 
$$
\frac{12!}{2!2!2!2!}
$$

For your second problem, keep in mind the the general binomial formula is
$$
(x+y)^n = \sum_{k=0}^{n}\binom{n}{k}x^{n-k}y^k.
$$
So here, you can simply take $x$ to be $3x$ (pardon the abuse of notation), and $y = -1$. Then the coefficient of $x^5$ is simply
$$
(3^5)(-1)^6\binom{n}{k}.
$$
