Difference between revisions of "2022 AMC 8 Problems/Problem 14"
MRENTHUSIASM (talk | contribs) (Created page with "==Problem== In how many ways can the letters in <math>\textbf{BEEKEEPER}</math> be rearranged so that two or more <math>\textbf{E}</math>s do not appear together? <math>\tex...") |
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==Problem== | ==Problem== | ||
| − | In how many ways can the letters in | + | In how many ways can the letters in BEEKEEPER be rearranged so that two or more Es do not appear together? |
<math>\textbf{(A) } 1 \qquad \textbf{(B) } 4 \qquad \textbf{(C) } 12 \qquad \textbf{(D) } 24 \qquad \textbf{(E) } 120</math> | <math>\textbf{(A) } 1 \qquad \textbf{(B) } 4 \qquad \textbf{(C) } 12 \qquad \textbf{(D) } 24 \qquad \textbf{(E) } 120</math> | ||
Revision as of 17:37, 31 January 2022
Problem
In how many ways can the letters in BEEKEEPER be rearranged so that two or more Es do not appear together?
Solution
All valid arrangements of the letters must be of the form ![\[\textbf{E\underline{\hspace{3mm}}E\underline{\hspace{3mm}}E\underline{\hspace{3mm}}E\underline{\hspace{3mm}}E}.\]](http://latex.artofproblemsolving.com/3/e/c/3ecf6e61439838584cdc342894a7e76208755b5c.png)
The problem is equivalent to counting the arrangements of 
and 
into the four blanks, in which there are 
ways.
~MRENTHUSIASM
See Also
| 2022 AMC 8 (Problems • Answer Key • Resources) | ||
| 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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