INTRODUCING 5 - days-a-week problem solving session for Math Olympiad and ISI Entrance. Learn More

Contents

[hide]

Try this beautiful problem from Combinatorics: Chosing Program

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?

,

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

Combinatorics

But try the problem first...

Answer: $9$

Source

Suggested Reading

AMC-10A (2013) Problem 7

Pre College Mathematics

First hint

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 ..........

Second Hint

**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........

Third Hint:

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

- https://www.cheenta.com/surface-area-of-cube-amc-10a-2007-problem-21/
- https://www.youtube.com/watch?v=VLyrlx2DWdA&t=20s

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.

JOIN TRIAL
Google