Difference between revisions of "2022 AMC 8 Problems/Problem 14"
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MRENTHUSIASM (talk | contribs) 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.) (Tag: Undo) |
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==Problem== | ==Problem== | ||
| − | In how many ways can the letters in BEEKEEPER be rearranged so that two or more | + | 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>\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 19:15, 31 January 2022
Problem
In how many ways can the letters in 
be rearranged so that two or more 
s 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
