Difference between revisions of "2013 AMC 10A Problems/Problem 7"
Imurmother (talk | contribs) (→Solution 2) |
Imurmother (talk | contribs) (→Solution 2) |
||
Line 21: | Line 21: | ||
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> | ||
− | |||
− | |||
− | |||
− | |||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2013|ab=A|num-b=6|num-a=8}} | {{AMC10 box|year=2013|ab=A|num-b=6|num-a=8}} |
Revision as of 02:15, 8 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 1
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 |