Difference between revisions of "2013 AMC 10A Problems/Problem 11"
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==Solution== | ==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 | + | 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>. |
== Solution 2 (much faster)== | == Solution 2 (much faster)== | ||
| − | 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}= | + | 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}} | ||
{{MAA Notice}} | {{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 
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?
Solution
Let the number of students on the council be 
. To select a two-person committee, we can select a "first person" and a "second person." There are choices to select a first person; subsequently, there are 
choices for the second person. This gives a preliminary count of 
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 
to account for overcounting. Thus, there are 
ways to choose the two-person committee. Solving this equation, we find that 
and 
are integer solutions. is a ridiculous situation, so there are people on the student council. The solution is 
.
Solution 2 (much faster)
To choose people from 
total people and get as a result, we can establish the equation 
which we can easily see 
, so there are people. The question asks how many ways to choose 
people from the , so there are 
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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