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Difference between revisions of "2013 AMC 10A Problems/Problem 11"
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==Problem==
 
==Problem==
  
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A student council must select a two-person welcoming committee and a three-person planning committee from among its members. There are exactly 10 ways to select a two-person team for the welcoming committee. It is possible for students to serve on both committees. In how many different ways can a three-person planning committee be selected?
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A student council must select a two-person welcoming committee and a three-person planning committee from among its members. There are exactly <math>10</math> ways to select a two-person team for the welcoming committee. It is possible for students to serve on both committees. In how many different ways can a three-person planning committee be selected?
  
  
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==Solution==
 
==Solution==
  
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Let the number of students on the council be <math>x</math>.  We know that there are <math>\dbinom{x}{2}</math> ways to choose a two person team. This gives that <math>x(x-1) = 20</math>, which has a positive integer solution of <math>5</math>.
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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 and so 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=10</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 solution is <math>\dbinom 5 3=10\implies \boxed{\textbf{A}}</math>.
  
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If there are <math>5</math> people on the welcoming committee, then there are <math>\dbinom{5}{3} = 10</math> ways to choose a three-person committee, <math>\textbf{(A)}</math>.
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== Solution 2 (much faster)==
  
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To choose <math>2</math> people from <math>n</math> total people and get <math>10</math> as a result, we can establish the equation <math>\binom{n}{2}=10</math> which we can easily see <math>n=5</math>, so there are <math>5</math> people. The question asks how many ways to choose <math>3</math> people from the <math>5</math>, so there are <math>\binom{5}{3}=\boxed{\textbf{(A)}10}</math> ways.
  
 
==See Also==   
 
==See Also==   
 
{{AMC10 box|year=2013|ab=A|num-b=10|num-a=12}}
 
{{AMC10 box|year=2013|ab=A|num-b=10|num-a=12}}
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{{MAA Notice}}

Latest revision as of 12:29, 27 July 2021

Problem

A student council must select a two-person welcoming committee and a three-person planning committee from among its members. There are exactly $10$ ways to select a two-person team for the welcoming committee. It is possible for students to serve on both committees. In how many different ways can a three-person planning committee be selected?


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

Solution

Let the number of students on the council be $x$. To select a two-person committee, we can select a "first person" and a "second person." There are $x$ choices to select a first person; subsequently, there are $x-1$ choices for the second person. This gives a preliminary count of $x(x-1)$ 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 and so latter and former arrangements would be considered the same. Therefore, we have to divide by $2$ to account for overcounting. Thus, there are $\dfrac{x(x-1)} 2=10$ ways to choose the two-person committee. Solving this equation, we find that $5$ and $-4$ are integer solutions. $-4$ is a ridiculous situation, so there are $5$ people on the student council. The solution is $\dbinom 5 3=10\implies \boxed{\textbf{A}}$.

Solution 2 (much faster)

To choose $2$ people from $n$ total people and get $10$ as a result, we can establish the equation $\binom{n}{2}=10$ which we can easily see $n=5$, so there are $5$ people. The question asks how many ways to choose $3$ people from the $5$, so there are $\binom{5}{3}=\boxed{\textbf{(A)}10}$ ways.

See Also

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

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