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

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~MRENTHUSIASM
 
~MRENTHUSIASM
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==Video Solution (CREATIVE THINKING!!!)==
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https://youtu.be/419vsFnrGeY
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~Education, the Study of Everything
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==Video Solution==
 
==Video Solution==
 
https://youtu.be/Ij9pAy6tQSg?t=1222
 
https://youtu.be/Ij9pAy6tQSg?t=1222

Revision as of 17:23, 28 March 2023

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 (CREATIVE THINKING!!!)

https://youtu.be/419vsFnrGeY

~Education, the Study of Everything

Video Solution

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

~Interstigation

Video Solution

https://youtu.be/p29Fe2dLGs8?t=212

~STEMbreezy

Video Solution

https://youtu.be/NmfnoSn3CDg

~savannahsolver

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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