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