The letters B, E, I, M, O, Z are arranged in alphabetical order. In a list where all the permutations are put, which place would the word ZOMBIE be? So I know that $6$ different letters can be arranged in $6!$ different ways. Which means
$$
6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720
$$
ways. and since the word $\mathsf{ZOMBIE}$ has the letters $\mathsf{Z}$, $\mathsf{O}$, $\mathsf{M}$ in the start means that $6 \times 5 \times 4$ disappears which leaves it $3!$ ways left to arrange it in. How should I go from here to find which place the word $\mathsf{ZOMBIE}$ is placed in out of the $720$ possible combinations?
How do I go from here to solve this?
 A: For each possible $1$st letter, there are $5! = 120$ possible ways to spell out the rest of the word. In other words, in alphabetical order, the first $120$ words begin with $\mathsf{B}$, the next $120$ words begin with $\mathsf{E}$, etc. So, the words that begin with $\mathsf{Z}$ are the final $120$ words, indexed from $5 \cdot 5! + 1 = 601$ through $5 \cdot 5! + 5! = 6! = 720$. In other words, we start with index $600$ and add whatever index $\mathsf{OMBIE}$ has among $5$-letter permutations.
Now, we remove the $\mathsf{Z}$ and consider each possible second letter. For each of the $5$ choices, there are $4! = 24$ possible ways to spell out the rest of the word, so that breaks up the range of indices into $5$ chunks, those starting with $\mathsf{ZB}$, those starting with $\mathsf{ZE}$, etc. The $5$th chunk of words begins with $\mathsf{ZO}$, so we need to add $4 \cdot 4! = 96$, so our running total is $600 + 96 = 696$. We will add to this whatever index $\mathsf{MBIE}$ has among $4$-letter permutations.
Continuing inductively, we can determine the index. Remember that the final letter in the permutation is completely determined once we choose the first $5$, so we must add $1$ at the end. Try to work this out yourself before revealing the spoiler.

 \begin{align} \color{purple}{5} &\cdot 5! + \color{blue}{4} \cdot 4! + \color{green}{3} \cdot 3! + \color{orange}{0} \cdot 2! + \color{red}{1} \cdot 1! + 1 \\ &= 600 + 96 + 18 + 0 + 1 + 1 \\ &= 716, \end{align} where the colored numbers count how many jumps (transpositions) are required to move each letter to its proper position in the word $\mathsf{ZOMBIE}$ from the alphabetical permutation $\mathsf{BEIMOZ}$, i.e. \begin{align} \color{purple}{\underline{\color{black} {\mathsf{BEIMO}}}}\mathsf{Z} \quad &\leadsto\quad \color{purple}{5} \\ \color{blue}{\underline{\color{black}{\mathsf{BEIM}}}}\mathsf{O} \quad &\leadsto\quad \color{blue}{4} \\ \color{green}{\underline{\color{black}{\mathsf{BEI}}}}\mathsf{M} \quad &\leadsto\quad \color{green}{3} \\ \mathsf{BEI} \quad &\leadsto\quad \color{orange}{0} \\ \color{red}{\underline{\color{black}{\mathsf{E}}}}\mathsf{I} \quad &\leadsto\quad \color{red}{1} \\ \end{align}

A: This is like counting in base $6$.
Let $B$, $E$, $I$, $M$, $O$ and $Z$ represent the digits $0\sim5$. Then we need to find the index of $543021$ among all numbers made by permuting $0\sim5$ each once ($0$ can be in lead), when they're listed in increasing order.
We just find how many numbers are bigger than $543021$, in this way it is easier. Their first three digits must be $5$, $4$ and $3$. So they're $543120$, $543102$, $543201$ and $543210$.
So its index is $6!-4=716$.
