Difference between revisions of "G285 MC10B Problems/Problem 1"

Latest revision as of 23:57, 3 June 2022

Find $\left \lceil {\frac{3!+4!+5!+6!}{2+3+4+5+6}} \right \rceil$

$\textbf{(A)}\ 42\qquad\textbf{(B)}\ 43\qquad\textbf{(C)}\ 44\qquad\textbf{(D)}\ 45\qquad\textbf{(E)}\ 46$

We have $\frac{6+24+120+720}{20} = \frac{87}{2} = \lceil 43.5 \rceil \implies \boxed{\textbf{(C)}\ 44}$

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+==Problem==
+Find <math>\left \lceil {\frac{3!+4!+5!+6!}{2+3+4+5+6}} \right \rceil</math>
+<math>\textbf{(A)}\ 42\qquad\textbf{(B)}\ 43\qquad\textbf{(C)}\ 44\qquad\textbf{(D)}\ 45\qquad\textbf{(E)}\ 46</math>
 ==Solution==
+We have <cmath>\frac{6+24+120+720}{20} = \frac{87}{2} = \lceil 43.5 \rceil \implies \boxed{\textbf{(C)}\ 44}</cmath>
-{{AMC12 box|year=2007|ab=A|num-b=24|after=Last question}}
-[[Category:Intermediate Combinatorics Problems]]
-{{MAA Notice}}