Difference between revisions of "Overcounting"
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* [[2004 AIME I Problems/Problem 3|AIME 2004I/3]] | * [[2004 AIME I Problems/Problem 3|AIME 2004I/3]] | ||
| + | ==Introductory Problems== | ||
| + | *How many different words can be formed with the letters <math>AAAABBCCDDDPPP</math>?(Not necessarily meaningful words) | ||
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[[Category:Definition]] | [[Category:Definition]] | ||
[[Category:Combinatorics]] | [[Category:Combinatorics]] | ||
Revision as of 10:36, 18 May 2012
Overcounting is the process of counting more than what you need and then systematically subtracting the parts which do not belong.
The Principle of Inclusion-Exclusion (PIE) is a systematic method of repeated overcounting that is a tool in solving many combinatorics problems.
An example of a classic problem is as follows:
"How many numbers less than or equal to 100 are divisible by either 2 or 3?"
Solution: Clearly, there are 50 numbers less than 100 that are divisible by 2, and 33 that are divisible by 3. However, we note that we overcount several numbers, such as 12, which is divisible by both 2 and 3. To correct for this overcounting, we must subtract out the numbers that are divisible by both 2 and 3, as we have counted them twice. A number that is divisible by both 2 and 3 must be divisible by 6, and there are 16 such numbers. Thus, there are 
numbers that are divisible by either 2 or 3.
(Note that it is not a coincidence that 67 is close to 2 thirds of 100! We can approach this problem in a constructive way, building the set based on the remainders when divided by 3, but that is a different subject).
Another basic example is combinations. In these, we correct for overcounting with division, by dividing out what we overcount (as opposed to above where we subtracted it out).
Here is MATHCOUNTS 2008 National Target #1: Try to solve this.
"How many numbers less than or equal to 100 are divisible by 2 or 3 but not 4?".
Now see we have another type of overcounting.For example, we have 
people in a party and everyone will handshake with each other.Suppose we are given a order to count the number of handshakes.Its not a matter of great deal if there are less than 10 persons.But assume a party from the President, there will be millions of people.Now the task seems impossible without mathematics.Suppose we know the total number of people invited in the party, say .
One person will handshake with all other except himself,i.e, he will handshake with 
people.Now since the total of people in the party is .So there will be 
handshakes.Now lets try our formula for two people.According to the formula we get 2 handshakes, but wait,we will have only 1 handshake between two persons.That means we have overcounted somewhere.
Actually, we counted 1 handshakes twice for the two persons.In our formula we have overcounted each handshake twice.One handshake for Person A to Person B and another for Person B to person A.So basically, 
where h is the number of handshakes.Now our formula tally with our experiment results.(To easily solve a problem, we overcount and then divide with number of times we overcounted)
Here is a question to try:
"How many diagonals does a n-sided polygon have?" Answer to questions and lots more is given in this (video) Other examples of overcounting are shown in the videos below.
Related Videos
- (Counting the Number of Arrangements of Letters in a Word)
- (Counting pairs)
- (Counting Objects in a Circle Part 1)
- (Counting Objects in a Circle Part 2)
Examples
Introductory Problems
- How many different words can be formed with the letters
?(Not necessarily meaningful words)
?(Not necessarily meaningful words)This article is a stub. Help us out by expanding it.
