2020 AMC 8 Problems/Problem 14

Revision as of 10:26, 28 January 2022 by Mahaler (talk | contribs) (Problem)

Problem

In how many ways can the letters in BEEKEEPER be rearranged so that two or more Es do not appear together?

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

Solution

For two or more Es to not appear together, the letters would have to be arranged in the form of E_E_E_E_E, where the blanks are one of each of the letters B, K, P, R. There are $4! = \boxed{24}$ ways you can arrange the letter in the blanks, so that is our answer.

~mahaler

Solution

We can see that the dotted line is exactly halfway between $4{,}500$ and $5{,}000$, so it is at $4{,}750$. As this is the average population of all $20$ cities, the total population is simply $4{,}750 \cdot 20 = \boxed{\textbf{(D) }95{,}000}$.

Video Solution by North America Math Contest Go Go Go

https://www.youtube.com/watch?v=IqoLKBx20dQ

~North America Math Contest Go Go Go

Video Solution by WhyMath

https://youtu.be/5y4uDwZEF0M

~savannahsolver

Video Solution

https://youtu.be/xjwDsaRE_Wo

Video Solution by Interstigation

https://youtu.be/YnwkBZTv5Fw?t=608

~Interstigation

See also

2020 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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All AJHSME/AMC 8 Problems and Solutions

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