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Difference between revisions of "2013 AMC 10A Problems/Problem 7"
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Let us split this up into two cases.
 
Let us split this up into two cases.
  
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Case 1: The student chooses both algebra and geometry.
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Case <math>1</math>: The student chooses both algebra and geometry.
  
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This means that 3 courses have already been chosen.  We have 3 more options for the last course, so there are <math>3</math> possibilities here.   
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This means that <math>3</math> courses have already been chosen.  We have <math>3</math> more options for the last course, so there are <math>3</math> possibilities here.   
  
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Case 2: The student chooses one or the other.   
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Case <math>2</math>: The student chooses one or the other.   
  
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Here, we simply count how many ways we can do one, multiply by 2, and then add to the previous.   
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Here, we simply count how many ways we can do one, multiply by <math>2</math>, and then add to the previous.   
  
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WLOG assume the mathematics course is algebra.  This means that we can choose 2 of History, Art, and Latin, which is simply <math>\dbinom{3}{2} = 3</math>.  If it is geometry, we have another 3 options, so we have a total of 6 options if only one mathematics course is chosen.
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WLOG assume the mathematics course is algebra.  This means that we can choose <math>2</math> of History, Art, and Latin, which is simply <math>\dbinom{3}{2} = 3</math>.  If it is geometry, we have another <math>3</math> options, so we have a total of <math>6</math> options if only one mathematics course is chosen.
  
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Thus, overall, we can choose a program in <math>6 + 3 = 9 ways</math>, $\textbf{(C)}
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Thus, overall, we can choose a program in <math>6 + 3 = 9</math> ways, <math>\textbf{(C)}</math>

Revision as of 15:34, 7 February 2013

A student must choose a program of four courses from a menu of courses consisting of English, Algebra, Geometry, History, Art, and Latin. This program must contain English and at least one mathematics course. In how many ways can this program be chosen?


$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 16$

Let us split this up into two cases.

Case $1$: The student chooses both algebra and geometry.

This means that $3$ courses have already been chosen. We have $3$ more options for the last course, so there are $3$ possibilities here.

Case $2$: The student chooses one or the other.

Here, we simply count how many ways we can do one, multiply by $2$, and then add to the previous.

WLOG assume the mathematics course is algebra. This means that we can choose $2$ of History, Art, and Latin, which is simply $\dbinom{3}{2} = 3$. If it is geometry, we have another $3$ options, so we have a total of $6$ options if only one mathematics course is chosen.

Thus, overall, we can choose a program in $6 + 3 = 9$ ways, $\textbf{(C)}$

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