Difference between revisions of "2013 AMC 10A Problems/Problem 11"

Revision as of 21:26, 10 November 2019

YEETING IS BANNED!!!

$\textbf{(A)}\ 10\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 25$

YEET SO MUCH THAT THEY HAVE TO KICK EVERYONE, OR ALLOW YEETING AGAIN!!!

2013 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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 AMC 10 Problems and Solutions

@@ Line 8: / Line 8: @@
 ==Solution==
-Let the number of students on the council be <math>x</math>.  To select a two-person committee, we can select a "first person" and a "second person." There are <math>x</math> choices to select a first person; subsequently, there are <math>x-1</math> choices for the second person. This gives a preliminary count of <math>x(x-1)</math> ways to choose a two-person committee. However, this accounts for the order of committees. To understand this, suppose that Alice and Bob are two students in the council. If we choose Alice and then Bob, that is the same as choosing Bob and then Alice; however in our counting, the latter and former arrangements would be considered the same. Therefore, we have to divide by <math>2</math> to account for overcounting. Thus, there are <math>\dfrac{x(x-1)} 2=20</math> ways to choose the two-person committee. Solving this equation, we find that <math>5</math> and <math>-4</math> are integer solutions. <math>-4</math> is a ridiculous situation, so there are <math>5</math> people on the student council. The sol is <math>\dbinom 5 3=10\implies \boxed{\textbf{A}}</math>.
+YEET SO MUCH THAT THEY HAVE TO KICK EVERYONE, OR ALLOW YEETING AGAIN!!!
 ==See Also==
 {{AMC10 box|year=2013|ab=A|num-b=10|num-a=12}}
 {{MAA Notice}}