Difference between revisions of "2013 AMC 10A Problems/Problem 7"
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===Solution 2=== | ===Solution 2=== | ||
| − | We can use complementary counting. Since there must be an English class, we will add that to our list of classes | + | We can use complementary counting. Since there must be an English class, we will add that to our list of classes leaving <math>3</math> remaining spots for rest of classes. We are also told that there needs to be at least one math class. This calls for complementary counting. The total number of ways of choosing <math>3</math> classes out of the <math>5</math> is <math>\binom{5}3</math>. The total number of ways of choosing only non-mathematical classes is <math>\binom{3}3</math>. Therefore the amount of ways you can pick classes with at least one math class is <math>\binom{5}3-\binom{3}3=10-1=\boxed{\textbf{(C) }9}</math> ways. |
===Solution 3=== | ===Solution 3=== | ||
| − | Similar to | + | Similar to Solution 1, note that for Case 1 of solution the answer is simply <math>\binom{4}2</math> and for the second case it is <math>\binom{3}2</math> hence <math>\boxed{\textbf{(C) }9}</math> ways |
| + | |||
| + | ==Video Solution (CREATIVE THINKING)== | ||
| + | https://youtu.be/ZGTHVQ6ixEo | ||
| + | |||
| + | ~Education, the Study of Everything | ||
| + | |||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/kDA-Yae2GPk | ||
| + | |||
| + | ~savannahsolver | ||
==See Also== | ==See Also== | ||
Latest revision as of 12:06, 1 July 2023
Contents
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
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.
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
Solution 2
We can use complementary counting. Since there must be an English class, we will add that to our list of classes leaving remaining spots for rest of classes. We are also told that there needs to be at least one math class. This calls for complementary counting. The total number of ways of choosing classes out of the 
is 
. The total number of ways of choosing only non-mathematical classes is 
. Therefore the amount of ways you can pick classes with at least one math class is 
ways.
Solution 3
Similar to Solution 1, note that for Case 1 of solution the answer is simply 
and for the second case it is 
hence 
ways
Video Solution (CREATIVE THINKING)
~Education, the Study of Everything
Video Solution
~savannahsolver
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
