Difference between revisions of "2013 AMC 10A Problems/Problem 7"
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Thus, overall, we can choose a program in <math>6 + 3 = 9</math> ways, <math>\textbf{(C)}</math> | Thus, overall, we can choose a program in <math>6 + 3 = 9</math> ways, <math>\textbf{(C)}</math> | ||
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| + | ==See Also== | ||
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| + | {{AMC10 box|year=2013|ab=A|num-b=6|num-a=8}} | ||
Revision as of 20:56, 7 February 2013
Problem
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?
Solution
Let us split this up into two cases.
Case 
: The student chooses both algebra and geometry.
This means that 
courses have already been chosen. We have more options for the last course, so there are possibilities here.
Case 
: The student chooses one or the other.
Here, we simply count how many ways we can do one, multiply by , and then add to the previous.
WLOG assume the mathematics course is algebra. This means that we can choose of History, Art, and Latin, which is simply 
. If it is geometry, we have another options, so we have a total of 
options if only one mathematics course is chosen.
Thus, overall, we can choose a program in 
ways, 
See Also
| 2013 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 6 |
Followed by Problem 8 | |
| 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 | ||
