# Verify my solutions to counting problems

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

1. 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 $\dfrac{12!}{2!2!2!2!}$.

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 $(1-x)^k = \ldots+\dbinom{k}kx^k$

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1. 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 $$\dfrac{12!}{2!2!2!2!}\;.$$

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 $\binom{11}5$.

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thanks, glad i cleared that up before the test – reddoor Mar 10 '13 at 22:38
@reddoor: You’re welcome. Yes, much better before than after! – Brian M. Scott Mar 10 '13 at 22:39
@reddoor: Note that the coefficient is actually $\binom{11}{5}(3^5)((-1)^8)$, but $(-1)^8=1$. If we wanted the coefficient in $(3x-1)^{12}$ the answer would be $-\binom{12}{5}(3^5)$. – André Nicolas Mar 10 '13 at 22:48
1. 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)

1. 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 $x^5$).

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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 $$(3^5)(-1)^6\binom{n}{k}.$$

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