Difference between revisions of "2022 AMC 8 Problems/Problem 14"

m (Undo revision 170799 by Pog (talk) Undo revision 170800 by Pog (talk) Let's keep the letters bolded I suppose. Bolded letters look really nice in my opinion.)
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~MRENTHUSIASM
 
~MRENTHUSIASM
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==Video Solution==
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https://youtu.be/Ij9pAy6tQSg?t=1222
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~Interstigation
  
 
==See Also==  
 
==See Also==  
 
{{AMC8 box|year=2022|num-b=13|num-a=15}}
 
{{AMC8 box|year=2022|num-b=13|num-a=15}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 23:46, 18 February 2022

Problem

In how many ways can the letters in $\textbf{BEEKEEPER}$ be rearranged so that two or more $\textbf{E}$s do not appear together?

$\textbf{(A) } 1 \qquad \textbf{(B) } 4 \qquad \textbf{(C) } 12 \qquad \textbf{(D) } 24 \qquad \textbf{(E) } 120$

Solution

All valid arrangements of the letters must be of the form \[\textbf{E\underline{\hspace{3mm}}E\underline{\hspace{3mm}}E\underline{\hspace{3mm}}E\underline{\hspace{3mm}}E}.\] The problem is equivalent to counting the arrangements of $\textbf{B},\textbf{K},\textbf{P},$ and $\textbf{R}$ into the four blanks, in which there are $4!=\boxed{\textbf{(D) } 24}$ ways.

~MRENTHUSIASM

Video Solution

https://youtu.be/Ij9pAy6tQSg?t=1222

~Interstigation

See Also

2022 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AJHSME/AMC 8 Problems and Solutions

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