Som sums' formula:

Pathfinder

From Mathcounts Mini :

Counting/Paths Along a Grid

From Art of Problem Solving

Counting Paths on a Grid 

Math Principles : Paths on a Grid : Two Approaches 

Question #1: How many ways to move the dominoes on a 6 by 6 checker board if you can only move the dominoes to the right or to the bottom starting from the upper left and you can't move the dominoes diagonally? 

Solution :
You can move the dominoes 5 times to the right at most and 5 down to
the bottom at most, so the answer is \(\dfrac {\left( 5+5\right) !} {5! \times 5!}\) = 252 ways

Question # 2: How many ways can you  move from A to B if you can only move downward and to right? 

Solution : There are \(\dfrac {\left( 4+4\right) !} {4!\times 4!}\) * 2 * \(\dfrac {\left( 4+4\right) !} {4!\times 4!}\) = 9800 ways from A to B

Sum of All the Possible Arrangements of Some Numbers

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Questions to ponder: (detailed solutions below) 

It's extremely important for you to spend some time pondering on these questions first without peeking on the solutions. 

#1: Camy made a list of every possible distinct four-digit positive integer that can be formed using each of the digits 1, 2 , 3 and 4 exactly once in each integer. What is the sum of the integers on Camy's list?

#2: Camy made a list of every possible distinct five-digit positive even integer that can be formed using each of the digits 1, 3, 4, 5 and 9 exactly once in each integer. What is the sum of the integers on Camy's list? (2004 Mathcounts Chapter Sprint #29)

#3: 2020 Mathcounts state sprint #24 













Solutions:
#1:  
Solution I: 
There are 4! = 24 ways to arrange the four digits. Since each digit appears evenly so each number will appear 24 / 4 = 6 times.
1 + 2 + 3 + 4 = 10 and 10 times 6 = 60 ; 60 (1000 +100 +10 + 1) = 60 x 1111 = 66660, which is the answer.

Solution II: 

The median of four numbers 1, 2, 3, 4 is (1 + 2 + 3 + 4) / 4 = 2.5 and there are 4! = 24 ways to arrange
the four numbers. 
2.5 (1000 + 100 + 10 + 1) x 24 = 66660 

#2:  
Solution I: 
Since this time Camy wants five-digit even integer, which means that the number "4" has to be at the unit digit and only 1, 3, 5, 9 can be moved freely.
Again there are 4! = 24 ways to arrange the four numbers. 1 + 3 + 5 + 9 = 18 and 18 x 6 = 108 (Each number that can be moved freely appears 6 times evenly.)108 x 11110 + 4 x 24 = 1199976

Solution II: 

Since this time Camy wants five-digit even integer, which means that the number "4" has to be at the unit digit and only 1, 3, 5, 9 can be moved freely. 
There will be 4! = 24 times the even number 4 will be used so 4 x 24 = 96
As for the remaining 4 numbers, their average (or mean) is \(\dfrac{1 + 3+ 5 + 9} {4} = 4.5\)                               
4.5 * ( 10000 + 1000 + 100 + 10) * 24 (arrangements)  + 96 = 4.5 * 11110 * 24 + 96 = 1199976

#3: The answer is 101. 

Other applicable problems: (answers below)

#1: What is the sum of all the four-digit positive integers that can be written with the digits 1, 2, 3, 4 if each digit must be used exactly once in each four-digit positive integer? (2003 Mathcounts Sprint #30)

#2: What is the average (mean) of all 5-digit numbers that can be formed by using each of the digits 1, 3, 5, 7, and 8 exactly once? (You can use a calculator for this question.) (2005 AMC-10 B)

#3: What is the sum of all the four-digit positive integers that can be written with the digits 2, 4, 6, 8 if each digit must be used exactly once in each four-digit positive integer?

#4: What is the sum of all the 5-digit positive odd integers that can be written with the digits 2, 4, 6, 8, and 3 if each digit must be used exactly once in each five-digit positive integer?  

#5:What is the sum of all the four-digit positive integers that can be written with the digits 2, 3, 4, 5 if each digit must be used exactly once in each four-digit positive integer?

  





Answer key: 
#1: 66660
#2: \(\dfrac{1 + 3 + 5 + 7 + 8}{5} = 4.8\)
4.8 * 11111 =\(\color{red}{53332.8}\) 
#3: 133320
#4: 1333272
#5: 93324

2015 Mathcounts State Prep: Mathcounts State Harder Questions

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Download this year's Mathcounts handbook here.

Question: 2010 Mathcounts State Team Round #10: A square and isosceles triangle of equal height are side-by-side, as shown, with both bases on the x-axis. The lower right vertex of the square and the lower left vertex of the triangle are at (10, 0). The side of the square and the base of the triangle on the x-axis each equal 10 units. A segment is drawn from the top left vertex of the square to the farthest vertex of the triangle, as shown. What is the area of the shaded region?

 There are lots of similar triangles for this question, but I think this is the fastest way to find the area.
  \(\Delta \)AGB is similar to \(\Delta \)DGC and their line ratio is 15 to 10 or 3 : 2.
     \(\Delta \)CGF is similar to \(\Delta \)CBE. 
     \(\dfrac {CG} {CB}=\dfrac {GF} {BE}\)
     \(\dfrac {2} {5}=\dfrac {GF} {10}\)\(\rightarrow GF=4\) From there you get the area =
      \(\dfrac{10\times 4} {2}=20\)

Question: 2010 Mathcounts State #30:  Point D lies on side AC of equilateral triangle ABC such that the measure of angle DBC is 45 degrees. What is the ratio of the area of triangle ADB to the area of triangle CDB? Express your answer as a common fraction in simplest radical form.

Since each side is the same for equilateral triangle ABC, once you use the 30-60-90 degree angle ratio and 45-45-90 degree angle ratio, you'll get the side.
Since area ratio stays constant, you can plug in any numbers and it's much easier to use integer first so I use 2 for \(\overline {CD}\).
From there you get the side length for each side is \(\sqrt {3}+1\).
\(\overline {AC}-C\overline {D}=\sqrt {3}+1-2\) = \(\overline {AD} = \sqrt {3}-1\)
\(\Delta ABD\) and \(\Delta CBD\) share the same vertex, so their area ratio is just the side ratio, which is \(\dfrac {\sqrt {3}-1} {2}\).

16 17 Mathcounts handbook more interesting questions that have nicer solutions

Thanks to Achuth for trying out these questions and time them as an actual Mathcounts test.  :) 

First week : warm up 1, 4, 7.  (time for 40 mins. like sprint)

Second week : warm up 2, 5, 8.

third week : workout 3 --> all right. (pair 1 to 6, 2 to 7, each time for 6 mins. as 

target) 

fourth week: workout 4 --> #95, then self correct. 

At lesson: workout 5 and other harder problems. 

These are nice questions that have various solutions, so it’s better to slow down and try them as puzzles.

Less is more and slow is fast.

If you are new to problem solving, one nice strategy is to make the question much simpler and explore ideas that come to your mind. 

Answer key down below. 

#66: A school of 100 fish swims in the ocean and comes to a very wide horizontal pipe. The fish have three choices to get to the food on the other side: swim above the pipe, through the pipe or below the pipe. If we do not consider the fish individually, in how many ways can the entire school of fish be partitioned into three groups with each group choosing a different one of the three options and with at least one fish in each group? 

 #105 When fully matured, a grape contains 80% water. After the drying process, called dehydration, the resulting raisin is only 20% water. What fraction of the original water in the grape remains after dehydration? Express your answer as a common fraction. 

 #112: Cora has five balls—two red, two blue and one yellow—which are indistinguishable except for their color. She has two containers—one red and one green. If the balls are randomly distributed between the two containers, what is the probability that the two red balls will be alone in the red container? Express your answer as a common fraction? 

 #116: A 12-foot by 12-foot square bathroom needs to be tiled with 1-foot square tiles. Two of the tiles are the wrong color. If the tiles are placed randomly, what is the probability that the two wrong-colored tiles share an edge? Express your answer as a common fraction.

#66: 4851

#105:  1/16

#112:  1/32

#116: 1/39