Counting I : Ways to Avoid Over Counting or Under Counting

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Question #1: How many ways to count a triple of natural numbers whose sum adds up to 12 
if  A. order doesn't matter? B. if order matters?

Solution I:
A. Find a systematic way to solve this type of problem in an organized manner.
We can start with the smallest natural number "1".
(1, 1, 10), (1, 2, 9), (1, 3, 8), (1, 4, 7), (1, 5, 6)  Since (1, 5, 6) is the same as (1, 6, 5) so stop.
(2, 2, 8), (2, 3, 7), (2, 4, 6), (2, 5, 5)
(3, 3, 6), (3, 4, 5)
(4, 4, 4) so total there are 12 ways.

B. Since in this case order matters so if you add up all the possible ways, for example, there are "3" ways to arrange (1, 1, 10) -- 3!/2! = 3
This is similar to "how many ways to arrange the letters 'odd'.
If all letters/numbers are different, you have 3! ways to arrange them; however, in this case, the two numbers 1, 1 are indistinguishable and there are 2! ways to arrange them, thus 3!/2! = 3

There are 3! or 6 ways to arrange triples such as (1, 2, 9).
Add all the possible ways and there are total 55 ways.

Solution II:
B.There is a much easier way to tackle this question, using bars and stars or sticks and stones method.
There are 11 spaces between 12 stones. @__@__@__@__@__@__@__@__@__@__@__@
If you places the two sticks on two of the spaces, you'll split the stones into three groups.
For example, if you have @ | @@@@@ | @@@@@@
The triples are 1, 5, and 6.
If it's @@@@@@@@ | @@ | @@ , the triples are 8, 2, and 2.
So 11C2 = 55 ways

Let me know if you are still confused.  Have fun problem solving !!

3 - D Geometry and harder AMCs, AIME links

For Mathcounts advanced group -- harder  Mathcounts State/National problem on 3-D geometry.                      

From Mathcounts Mini :

Using Similarity to Solve Geometry Problems

Area and Volume 

Relationship Between A Box's Dimensions, Volume and Surface Area

AMC, AIME videos from AoPS

Harder/Hardest AMC-10, 12 and AIME problems explained by Mrs. Rusczyk

Weird but Delicious Math Questions

Check out Mathcounts here--the best competition math for middle school mathletes.

Problem: (Solution below)
#1: 1993 Mathcounts National Team Round #4 :The teacher whispers positive integer A to Anna, B to Brett, and C to Chris. The students don't know one another's numbers but they do know that the sum of their numbers is 14. Anna says: "I know that Brett and Chris have different numbers." Then Brett says: "I already knew that all three of our numbers were different." Finally, Chris announces: "Now I know all three of our numbers." What is the product ABC?

#2: 2000 AMC10 #22: One morning each member of Angela’s family drank an 8-ounce mixture of coffee with milk. The amounts of coffee and milk varied from cup to cup, but were never zero. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. How many people are in the family?

Solution: 

#1: For Anna to know that Brett and Chris have different numbers, she must have an odd number
because 14 - odd = odd and you can only get odd sum of two numbers if they are different, one odd
one even.

For Brett to say that he already know all three have different numbers, he not only must have an odd number, but also his number has to be larger or equal to 7 and is not the same as what A have.

Otherwise, A-B-C could be 1-1-12; 3-3-8; or 5-5-4.
It would exceed 14 if you have 7-7-__. (All numbers are positive) so Brett"s and Anna's numbers must
be different.

If Brett has 7, then the numbers could be A-B-C = 1-7-6 ; 3-7-4 or 5-7-2.

If Brett has 9, then the numbers could be A-B-C = 1-9-4; or 3-9-2.

If Brett has 11, then then numbers could be A-B-C = 1-11-2.

From the above possibilities you know Chris has to have 6 for him to be sure he knows all the numbers. 

So A-B-C = 1-7-6 and the product of ABC = 1 x 7 x 6 = 42

#2: Let there be m cups of mild, c cups of coffee. and x people in Angela's family.
According to the given, you can set up the following equation: 
  m + c = = 8. Two ways to solve this equation.

Solution I:

m + c =   Both sides times 12 x

Both m and c need to be positive so the only x that works is when x = 5.

Solution II: 

m + c  = 8, Both sides times 12 and you have 

3m + 2c = 96; m + c is a multiple of 8.

30      3

28      6

26      9

.

.

.

2       45;  from  (30 + 3 ) = 33 to (2 +45) = 47 only 40 is a multiple of 8

and 40 divided by 8 = 5 so 5 is the answer.