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AMC 10 Geometry Math Olympiad USA Math Olympiad

Chosing Program | AMC 10A, 2013 | Problem 7

Try this beautiful problem from Combinatorics based on Chosing Program from AMC-10A (2013), Problem 7. You may use sequential hints to solve the problem.

Try this beautiful problem from Combinatorics: Chosing Program

Chosing Program – AMC-10A, 2013- Problem 7


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?

,

  • $6$
  • $8$
  • $9$
  • $12$
  • \(16\)

Key Concepts


Combinatorics

Check the Answer


Answer: $9$

AMC-10A (2013) Problem 7

Pre College Mathematics

Try with Hints


There are six programms: English, Algebra, Geometry, History, Art, and Latin. Since the student must choose a program of four course with the condition that there must contain English and at least one mathematics course. Therefore one course( i.e English) are already fixed and we have to find out the other subjects combinations…….

Can you now finish the problem ……….

There are 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.

Let us choose the mathematics course is algebra. so we can choose 2 of History, Art, and Latin, which is simply $3 \choose 2$=$3$. If it is geometry, we have another 3 options, so we have a total of 6 options if only one mathematics course is chosen.

can you finish the problem……..

Therefore the require ways are \(6+3=9\)

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