2021 WSMO Speed Round Problems/Problem 5

Problem

The number of ways to arrange the characters in "delicious greenbeans" into two separate strings of letters can be expressed as $a\cdot b!,$ where $b$ is maximized and both $a$ and $b$ are positive integers. Find $a+b.$ (A string of letters is defined as a group of consecutive letters with no spaces between them.)

Solution

Note that "delicious greenbeans" has 19 letters. In addition, the "delicious greenbeans" has 1 d, 4 e's, 1 l, 2 i's, 1 c, 1 o, 1 u, 1 g, 1 r, 2 s's, 2 n's, 1 b, and 1 a. Thus, there are $\frac{19!}{1!4!1!2!1!1!1!1!@!2!1!1!}=\frac{19!}{4!2!2!2!}$ distinct ways to arrange the letters in "delicious greenbeans". In addition, since there are 19 letters, there are 18 possible places to put the space (" "). Thus, the answer is \[\frac{19!}{4!2!2!2!}\cdot18=\frac{19!}{32}\cdot3=\frac{3\cdot19\cdot18\cdot17\cdot16\cdot15!}{32}=3\cdot19\cdot9\cdot17\cdot15!=8721\cdot15!\Longrightarrow8721+15=\boxed{8736}\]

~pinkpig