Difference between revisions of "2020 AMC 8 Problems/Problem 14"
MRENTHUSIASM (talk | contribs) m (→Solution 1) |
(→Problem) |
||
Line 1: | Line 1: | ||
==Problem== | ==Problem== | ||
− | |||
− | + | 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\qquad</math> | |
− | + | ==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 <math>4! = \boxed{24}</math> ways you can arrange the letter in the blanks, so that is our answer. | |
− | |||
− | |||
− | |||
− | + | ~mahaler | |
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
==Solution== | ==Solution== |
Revision as of 10:26, 28 January 2022
Contents
Problem
In how many ways can the letters in BEEKEEPER be rearranged so that two or more Es do not appear together?
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 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 and
, so it is at
. As this is the average population of all
cities, the total population is simply
.
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
~savannahsolver
Video Solution
Video Solution by Interstigation
https://youtu.be/YnwkBZTv5Fw?t=608
~Interstigation
See also
2020 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.