Part I work : progress report sent through e-mail.
Part II work : online timed test of trickier problems.
Mathcounts mini on 30-60-90 degree special right triangles.
Notes on problems you got wrong the most :
#1: When the problem asks about the probability of getting even product, complementary counting is usually faster.
#2: Counting problems are tricky, especially those that involve cases so do it later, but most importantly, learn to count systematically and get organized.
Practice this : probability that when tossing two dice the product is larger than 10 is
17/36
#3: This 2013 AMC-10 problem stumps quite a lot of you. It's not hard so make sure you understand how to approach it.
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 I:
Case 1: With two math courses -- English, Algebra, Geometry so only one course is left to choose from the remaining
three, and 3C1 = 3 ways.
Case 2: With just one math courses -- English , (Algebra or Geometry -- two ways), once two courses are set, two remaining courses are left to choose from the other 3 courses (History, Art or Latin).
Thus 2 (Algebra or Geometry) * 3C2 (from History, Art or Latin) = 6
Add them up and the answer is 9. -- Nishant got it right, using this method.
Solution II:
Use complementary counting : Besides English, you need to take 3 courses from the other 5 courses offered.
Thus 5C3 = 10. However, you need to take at least one math course so minus the case that involves "no math
courses, which is 3C3 = 1 (History, Art and Latin)
5C3 - 3C3 =9 ways. -- Ashna got it right, using this method.
From Mathcounts Mini :
Case work Counting part 1
Case work Counting part 2
Counting pairs
Probability and equally likely event
Part II work : online timed test of trickier problems.
Mathcounts mini on 30-60-90 degree special right triangles.
Notes on problems you got wrong the most :
#1: When the problem asks about the probability of getting even product, complementary counting is usually faster.
#2: Counting problems are tricky, especially those that involve cases so do it later, but most importantly, learn to count systematically and get organized.
Practice this : probability that when tossing two dice the product is larger than 10 is
17/36
#3: This 2013 AMC-10 problem stumps quite a lot of you. It's not hard so make sure you understand how to approach it.
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 I:
Case 1: With two math courses -- English, Algebra, Geometry so only one course is left to choose from the remaining
three, and 3C1 = 3 ways.
Case 2: With just one math courses -- English , (Algebra or Geometry -- two ways), once two courses are set, two remaining courses are left to choose from the other 3 courses (History, Art or Latin).
Thus 2 (Algebra or Geometry) * 3C2 (from History, Art or Latin) = 6
Add them up and the answer is 9. -- Nishant got it right, using this method.
Solution II:
Use complementary counting : Besides English, you need to take 3 courses from the other 5 courses offered.
Thus 5C3 = 10. However, you need to take at least one math course so minus the case that involves "no math
courses, which is 3C3 = 1 (History, Art and Latin)
5C3 - 3C3 =9 ways. -- Ashna got it right, using this method.
From Mathcounts Mini :
Case work Counting part 1
Case work Counting part 2
Counting pairs
Probability and equally likely event