2017 Mathcounts State Prep: Some Counting and Probability Questions on Dot Grids

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#5 1993 Mathcounts National Target : Find the probability that four randomly selected points on the geoboard below will be the vertices of a square? Express your answer as a common fraction.

#5 2004 AMC 10A: A set of three points is chosen randomly from the grid shown. Each three-point (same image as the below question) set has the same probability of being chosen. What is the probability that the points lie on the same straight line?

2007 Mathcounts Chapter Sprint #29 : The points of this 3-by-3 grid are equally spaced
horizontally and vertically. How many different sets of three points of this grid can be the three
vertices of an isosceles triangle?

Solution:

#5 National Target: There are 16C4 = \(\dfrac {16\times 15\times 14\times 13} {4\times 3\times 2\times 1}\)= 1820 ways to select 4 points on the geoboard.

There are 3 x 3 = 9  one by one squares and 2 x 2 = 4 two by two squares and 1 x 1 = 1 three by three squares. (Do you see the pattern?)

                                                         

There are 4 other squares that have side length of √ 2
and 2 other larger squares that have side length of √ 5.

9 + 4 + 1 + 4 + 2 = 20 and \(\dfrac {20} {1820}=\dfrac {1} {91}\)

#5: Solution:
AMC-10A: There are 9C3 = \(\dfrac {9\times 8\times 7} {3\times 2\times 1}\)= 84 ways to chose the three dots and 8 of the lines connecting the three dots will form straight lines. (Three verticals, three horizontals and two diagonals.) so 

\(\dfrac {8} {84}=\dfrac {2} {21}\)

#29: Solution: 
Use the length of the two congruent legs to solve this problem systematically. 

 There are 16   1 - 1 - \(\sqrt {2}\)    isosceles triangles.
There are 8    \(\sqrt {2}\)  by  \(\sqrt {2}\) by 2 isosceles triangles. (See that ?)

There are 4     2 - 2 - \(2\sqrt {2}\)  isosceles triangles.
There are 4    \(\sqrt {5}\)  by  \(\sqrt {5}\) by 2 isosceles triangles.
Finally, there are 4  \(\sqrt {5}\)  by  \(\sqrt {5}\) by \(\sqrt {2}\) isosceles triangles.
Add them up and the answer is 36.  

2013 Mathcounts State Prep: Counting Problems

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#1: 2006 Mathcounts state : My three-digit code is 023. Reckha can’t choose a code that is the same as mine in two or more of the three digit-positions, nor that is the same as mine except for switching the positions of two digits (so 320 and 203, for example, are forbidden, but 302 is fine). Reckha can otherwise choose any three-digit code where each digit is in the set {0, 1, 2, ..., 9}. How many codes are available for Reckha?

Solution:
Do complementary counting. Use total possible ways minus those that are not allowed. 
 
You can't use two or more of the numbers that are at the same position (given) as 203, which means that you can't have 0 __ 3, __ 23, or 02__.

For each of the __, you can use 10 digits (from 0, 1, 2 ... to 9) so 10 + 10 + 10 = 30.

However, you repeat 023 three times in each case so you need to minus 2 back so not to over count.
30-2 = 28

Also, you can't just switch two digits, which means 320, 203 and 032 are not allowed.  { but 302 and 230 are allowed since you are switching all the digits }

There are 10 x 10 x 10 = 1000 digits total and 1000 - 28 - 3 = 969   The answer

#2: 2011 AMC-8 # 23: How many 4-digit positive integers have four different digits where the leading digit is not zero, the integer is a multiple of 5, and 5 is the largest digit?

Solution: 
For the integer to be a multiple of 5, there are two cases: 

Case I: The unit digit is 5 :  __ __ __ 5

There are 4 numbers to choose for the thousandth digit [since 5 is the largest digit and you can't have "0" for the leading digit so there are 4 numbers 1, 2, 3, 4 that you can use], 4 numbers to choose for the hundredth

digit (0 and one of the remaining 3 numbers that are not the same number as the one in the thousandth digit) and 3 numbers to choose for the tenth digit (the remaining 3 numbers) so total 4 x 4 x 3 = 48 ways

Case II:  The unit digit is 0: __ __ __ 0

One of the remaining three numbers has to be 5, and for the remaining 2 numbers, there are 4C2 = 6 ways

to choose the 2 numbers from the numbers 1, 2, 3 or 4.
There are 3! arrangements for the three numbers so 6 x 3! = 36

48 + 36 = 84 ways

2013 Mathcounts State Prep: Partition Questions

#24 2001 Mathcounts Sate Sprint Round: The number 4 can be written as a sum of one or more natural numbers in exactly five ways: 4, 3+1, 2 + 1 + 1, 2 +2 and 1 + 1 + 1 + 1; and so 4 is said to have five partitions. What is the number of partitions for the number 7?

#2: Extra: Try partition the number 5 and the number 8. 

Solution: 

#24: You can solve this problem using the same technique as counting coins:

Counting coins questions

7     6    5    4    3    2    1

1                                           1 way

       1                                    1 way

             1                 1           2 ways ( 5 + 2 or 5 + 1 + 1)

                   1    1                  1 way

                   1           1           2 ways ( 4 + 2 + 1 or 4 + 1 + 1 + 1)

                         2     0   1      1 ways

                         1     2           3 ways ( 3 + 2 + 2, 3 + 2 + 1 + 1 and 3 + 1 + 1 + 1 + 1)

                                3           4 ways (2 + 2 + 2 + 1, 2 + 2 + 3 ones, 2 + 5 ones and 7 ones.)

Total 15 ways.

The partitions of 5 are listed below (There are 7 ways total.):

5   4   3   2   1

1                           1 way

     1                      1 way

          1    1           2 ways  (3 + 2 and 3 + 1 + 1)

                2           3 ways  (2 + 2 + 1, 2 + 1 + 1 + 1 and 1 + 1 + 1 + 1 + 1)

There are 22 ways to partition the number 8.

How Many Numbers (Terms)? Space, Inclusive and Exclusive Notes

Please write a comment and give me feedback. Thanks a lot!!

Quite a lot of my students have problems figuring out this type of problems so here are the notes.

#1: How many consecutive numbers from 1 to 5 inclusive?

1 _ 2 _ 3 _ 4 _ 5   There are 5 numbers if you just list them and count them out; however, what if
the question is:

#2: How many consecutive numbers from 34 to 200 (SAT type problem)?

Most students would think it's 200 - 34 = 166, but it's not.

Using #1 case, if you do 5 - 1 = 4, you are only getting how many spaces between those consecutive numbers.

Thus for question #2, the correct answer is 200 - 34 + 1 or 200 - 33 = 167

#3: What about how many consecutive numbers from 5 to 100 exclusive?

Inclusive means including the first and the last numbers; exclusive means not including the first and the last numbers, so for this question, you do 100 - 5 - 1 = 94.

Use # 1 case to help you figure out and really understand the concepts involved.

Here are other questions to help you practice the skills.

Word problems:  Answers below.

#1: How many numbers from 45 to 100 inclusive? 

#2: How many numbers from 17 to 127 inclusive? 

#3: How many numbers from 12 to 34 exclusive? 

#4: How many multiples of 9 from 1 to 200 inclusive?

#5: The dimension of the square on the left is 20 feet by 20 feet. If you  place a post every four feet, starting at one corner, how many posts will be placed? 

#6:  The distance from exit 13 to 21 is 216 miles. How many miles is the distance between two exits if all exits are equally spaced? 

#7: How many multiples of 5 from 120 to 218 exclusive? 

#8: Who is right? The teacher or the student?  Try this question.

#9: How many numbers from -12, -11, -10.........56 inclusive? 

What is their sum?

#10: How many numbers are in the list: 17.25, 18.25, 19, 25...111.25? 

Solutions: To excel at Mathcounts state/national, you need to practice all these questions mentally.

#1: 100 - 45 + 1 = 100 - 44 = 56
#2: 127-17 + 1 = 127 - 16 = 111
#3: Exclusive: 34 -12 -1 = 34 - 13 = 21
#4: Multiples of 9 from 1 to 200 starts with 9 and ends in 198.

Solution I: 9 , 18, 27...198  = 9 (1, 2, 3, ...22) The answer is 22.

Solution II:   \(\frac{(198 - 9)}{9} + 1 = 22"\)

#5:  Just observe one side first. Exclude the 4 corners, the other posts
are similar to those exclusive type problems.
There are posts on each side so 4 * 4 + 4 (corner posts)
= 20



 #6: There are 21 - 13 = 8 space so miles.  The answer is 27 miles.
#7:  Multiples of 5 from 120 to 218 start with 120 and end in 215.
Since it's asking exclusive, 120, 125, ...215 = 5(24, 25, ...43)
43 - 24 - 1 = 43 - 25 = 18
#8: You only need two cuts to get 3 pieces so 2 * 10 = 20 minutes. The student is right.
#9: 56 - ( -12) + 1 = 56 + 13 = 69
The sum is from 13 to 56 since up to 12 it got cancelled with the negative equivalent numbers.
Use average * the term you got  . The sum is 1518.
#10:111.25 - 117.25 + 1 = 111.25 - 116.25 = 95

Problem Solving Strategy: Counting Coins

Q: How many ways can you make 25 cents if you can use quarters, dimes, nickels or pennies?
Start with the largest value:

1 Quarter      1 way

2 Dimes, 1 Nickel   You can stop here since it implies 2 ways.
2 Dimes, 0 Nickel ( which implies 5 pennies)

1 Dime, 3 Nickels , which implies 4 ways.
1 Dime, 2 Nickels ( 5 pennies)
1 Dime, 1 Nickel   (10 pennies)
1 Dime, 0 Nickel ( 15 pennies)

0 Dime, 5 N, which implies 6 ways.
: , 4 N  (5 pennies)
: , 3 N  (10 pennies)
: , 2 N  (15 pennies)
: , 1 N  (20 pennies)
: , 0 N (25 pennies)

So altogether 13 ways.

Practice questions: (answers below)

Q 1 : How many ways can you make a. 15cents, b. 20cents, c. 30 cents if you can use quarters, dimes, nickels or pennies?

Q 2 : How many different combination of coins could a person have if she has exactly 21 cents?

Q 3 :Using nickels, dimes, quarters and/or half-dollars, how many ways can you make 75 cents?

Q 4: 20 coins of quarters and nickels add up to 4 dollars. How many nickels are there? 

Q5:  What is the least number of US coins to make changes possible from 1 to  99 cents inclusive? (half dollar is allowed)










Answers: 

#1: a. 6 ways ; b. 9 ways; c 18 ways ;
#2: 9 ways
#3: 22  ways
#4: 5 nickels and 15 quarters   
#5: 9 coins (1 half dollar, 1 quarter, 2 dimes, 1 nickel and 4 pennies)