Download this year's Mathcounts handbook here.

Video to watch on complementary counting from "Art of Problem Solving"

Part 1

Part 2

Question:

**How many two-digit numbers contain at least one 9?**

**At beginning level, kids start to write down all the numbers that contain 9. However, this turns into impossible task if it's a three-digit or four-digit number. So let's try other ways to do it.**

9 _ , if 9 is placed as the tens digit, unit digit can be chosen from 0 -9, altogether 10 numbers.

A lot of kids think there are only 9. Be careful. To find how many consecutive terms from number a to b, you do (b-a) +1

_ 9, there will be 9 choices (1-9) this time as the tens digit.(Why?) Zero can't be placed other than unit digit.

So total you have 10 + 9 - 1 (you've counted 99 twice) = 18

This way is better than the first one, but once the numbers become large, you will easily lose track of those double-counting, triple-counting numbers and over count your answers.

Here is a better way to tackle this type of problem: Think about the case of numbers that contain no 9s, and subtracting this value from the total number of two-digit numbers will give you the answer.

There are 99-10 +1 =90 total two digit numbers. There are 8 (the tens digit) x 9 (unit digit) = 72 numbers that contain no 9s. So 90-72 = 18 gives you the answer.

Try this question:

**How many three-digit numbers contain at least one 9?**

There are 999 - 100 + 1 or 999 - 99 = 900 three-digit numbers.

Or 9 x 10 (you can use the digit "0" now" x 10 = 900 three-digit numbers

900 - 8 x 9 x 9 =

**252**

**numbers**

This is called

**"complementary counting"**and there are numerous problems that you can use this strategies to simplify the reasoning.

Here is another harder problem from 2003 Mathcounts Chapter Sprint Round #29:

**Each day, two out of the three teams in a class are randomly selected to participate in a MATHCOUNTS trial competition. What is the probability that Team A is selected on at least two of the next three days? Express your answer as a common fraction.**

**Solution:**

Use complementary counting.

If each day two of the team will be chosen, there will be 3C2 = 3 ways to choose the team -- AB, BC, or AC, so 1/3 of the chance that team A won't be chosen and 2/3 of the chance that team A will be chosen.

Case 1: Team A is not chosen on any of the three days. The probability is (1/3)

^{3}= 1/27.

Case 2: Team A is chosen on one of the three days : The probability is (2/3) times (1/3)

^{2}times 3C1 = 6/27 (A - -, - A - or - - A, which is 3C1 = 3 ways)

Total possibilities - none - at least 1 time = at least two times Team A will be chosen

so the answer is 1 - 1/27 - 6/27 =

**20/27**

**Other applicable problems: (answer key below)**

**#1: 2006 AMC10 A: How many four-digit positive integers have at least one digit that is a 2 or 3?**

**#2: What is the probability that when tossing two dice, at least one dice will come up a "3"?**

**#3: If {x,y} is a subset of S={1,2,3,....50}. What is the probability that xy is even?**

**Answer key:**

**#1: 9000 - 7 x 8 x 8 x 8 =**

**5416**

**#2: The probability of the dice not coming up with a "3" is 5/6.**

1 - (5/6)

^{2}=

**11/36**

**2. 1 - (25/50) (24/49) -- only odd times odd will give you odd product, the others will all render even product, so the answer is**

**37/49**.