Difference between revisions of "2021 WSMO Speed Round Problems/Problem 5"
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==Solution== | ==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 <math>\frac{19!}{1!4!1!2!1!1!1!1!@!2!1!1!}=\frac{19!}{4!2!2!2!}</math> 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 <cmath>\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}</cmath> | 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 <math>\frac{19!}{1!4!1!2!1!1!1!1!@!2!1!1!}=\frac{19!}{4!2!2!2!}</math> 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 <cmath>\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}</cmath> | ||
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Latest revision as of 10:20, 23 December 2021
Problem
The number of ways to arrange the characters in "delicious greenbeans" into two separate strings of letters can be expressed as 
where 
is maximized and both 
and are positive integers. Find 
(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 
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 ![\[\frac{19!}{4!2!2!2!}\cdot18=\frac{19!}{32}\cdot3=\frac{3\cdot19\cdot18\cdot17\cdot16\cdot15!}{32}=3\cdot19\cdot9\cdot17\cdot15!=8721\cdot15!\Longrightarrow8721+15=\boxed{8736}\]](http://latex.artofproblemsolving.com/e/d/6/ed6f086903b2eb882da02d604db0fac59077cdcb.png)
~pinkpig
