How many $7$- letter words can be formed from the letters in bookkeeper? I was thinking that I would write it out like $\dfrac {10} 3 \dfrac 8 2 \dfrac 6 2 5!$, but I realized that would just show the total possible ways the word can be rearranged. If I were to guess as to what to do I would say that $\dfrac {10} 7$, which translates to $\dfrac {10!} {7!} 3!$.
BOOKKEEPER: one each B,P,R, two each K,O, and three E.

To form a seven letter word (assuming not limited to actual English words) we may select


a triple and two doubles: $\binom{1}{1}\binom{2}{2}\frac{7!}{3!2!^2}$
a triple, double, and two singles: $\binom{1}{1}\binom{2}{1}\binom{4}{2}\frac{7!}{3!\,2!\,1!}$
a triple, four singles: $\binom{1}{1}\binom{5}{4}\frac{7!}{3!\,1!^4}$
three doubles, one single: $\binom{3}{3}\binom{3}{1}\frac{7!}{2!^3\,1!}$
two doubles, three singles: $\binom{3}{2}\binom{4}{3}\frac{7!}{2!^2\,1!^3}$
one double, five singles $\binom{3}{1}\binom{5}{5}\frac{7!}{2!\,1!^5}$
seven singles: $\binom{6}{7}\frac{7!}{1!^7}$  (whut?)


$$\frac{7!}{3!2!^2}+\frac{7!\,12}{3!\,2!}+\frac{7!\,5}{3!}+\frac{7!\,3}{2!^3}+\frac{7!\,12}{2!^2}+\frac{7!\,3}{2!}+0$$
you have a multiset {${1\cdot B, 3\cdot E, 2\cdot K, 2\cdot O, 1\cdot R, 1\cdot P}$} There are 6 letters in this set.

First look at taking one of each letter on then one double (E,K or O) this can be done in $3\choose 1$ ways and the chosen letters can be arranged in $\frac{7!}{2!}$ ways.This give a total of (${3\choose 1}\cdot \frac{7!}{2!}$) ways. Follow on with this by looking at taking 2 doubles, 3 doubles, a double and a triple and 2 doubles and a triple to get your final answer. Note the only possible triple is 'E'. All of these are straight forward with the exception of 2 doubles which will require some thought.