2008 AMC 8 Problems/Problem 14

Revision as of 04:10, 7 August 2025 by Sinchana (talk | contribs) (Problem)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Solution

There are $2$ ways to place the remaining $\text{As}$, $2$ ways to place the remaining $\text{Bs}$, and $1$ way to place the remaining $\text{Cs}$ for a total of $(2)(2)(1) = \boxed{\textbf{(C)}\ 4}$.

Video Solution

https://www.youtube.com/watch?v=8qzMymleTIg ~David

Video Solution 2

https://youtu.be/1m_c_iMvxKo Soo, DRMS, NM

See Also

2008 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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png