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By Patrick, Andrew, Jared & Jonathan

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 Counting is the study of the enumeration of discrete, finite sets (“counting objects”)

 Ex. If there are 5 symbols with which we are given, to create a 4 symbol license plate, how many plates can we make? (Symbols can be repeated)  (5)(5)(5)(5) = 625

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ULTIPLICATION

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RINCIPLE FOR

 If an activity can be performed as a sequence of “k” independent steps, and step “i” can be accomplished in “n i ” ways, than the entire activity can be performed in (n 1 )(n 2 )(n 3 )…(n k ) ways.

 When can this be used?

 Ex. If we must create a license plate in which two of the symbols can be of a given set of 6 letters or two of the symbols can be of a given set of 5 numbers, how many unique plates can we make?

 (5)(5) + (6)(6) = 61

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DDITION

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RINCIPLE FOR

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 i If an activity can be divided into “k” disjoint subsets of activities (separate activities), and the th of these sets contains n i number is n 1 + n 2 + n 3 +….+ n k .

elements, than total

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ERMUTATION

   A permutation of “n” distinct objects x 1 , x 2 , x 3 …x n is any ordering of “n” objects.

P(n,r) = n!/(n-r)! where one chooses “r” items from “n” items (order matters).

Where does it come from?

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XAMPLE

   If there are 16 baseball players who need to be put into a 3 player batting order, how many different batting orders are possible?

P(16,3) = 16!/(16-3)! = 16!/13! = 3,360.

There are 3,360 unique batting orders which are possible.

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OMBINATIONS

  Combinations are Permutations where the order of the objects being counted doesn’t matter.

For example: 1, 2, and 3 can be arranged in 6 different ways if order matters but only 1 way if order does matter: Matters Doesn’t Matter 1,2,3 1,3,2 2,1,3 2,3,1 3,2,1 3,1,2 1,2,3

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OMBINATIONS

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 The table from the previous page shows that for a set of (1,2,3) if you want to order the numbers with a Permutation you will get 6 outputs, but with a Combination, you only receive 1.

 This shows that there are 6 times as many Permutations than combinations for this set. This works for any P(n,r) and C(n,r).

 Since there are 3 digits, you divide the number of permutations by another 3! which is equal to 6.

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OMBINATIONS

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QUATION

 To reach this conclusion we reduce the formula by how many ways the objects could be in order which accounts for the r! on the bottom.

 P(n,r) = C(n,r)*r!

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OMBINATIONS WITH REPETITION

 Combinations with repetition are also referred to as “Multichoose” problems.

 The equation is  How does it work?

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ULTICHOOSE

 Let’s say you have 5 different flavors of ice cream, and you can have 3 scoops. Your 3 scoops can be any flavor, with repetition, how many different combinations of ice cream scoops can you have?

 We will say the flavors are banana, chocolate, lemon, strawberry and vanilla.

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ULTICHOOSE

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 Think about the ice cream being in containers, you could say "skip the first, then 3 scoops, then skip the next 3 containers" and you will end up with 3 scoops of chocolate!  So there are r + (n-1) positions, and we want to choose r of them to have circles.

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XAMPLE

  If we refer back to the previous problem about batting orders, we will take the same number of players, 16, but this time 3 players will be picked to be team captains. How many ways can they be chosen in groups of 3?

C(16,3) = 16!/3!(16-3)!

 = 16!/3!(13)!

= 16*15*14/3*2*1 = 560 In this case, there are only 560 ways that players can be chosen, because order does not matter.

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DENTITY

130   Identity: For 0 ≤ k ≤ n

k( n k )=n( n-1 k-1 )

Question: How many ways can we create a size k committee of students from a class of n students, where one of the committee members is designated as the chair?

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DENTITY

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CONTINUED

  There are (

n k

) ways to choose the committee, then k ways to select a chair. Therefore, there are k( n k ) possible outcomes.

First, select the chair from the class of n students. Then, from the remaining n-1 students, pick the remaining k-1 members. This can be done n( n-1 k-1 ) ways.

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DENTITY

131  We have a question we could ask:  How many ways can we make a committee (of any size) from a class of students of n students, where one of them is designated as the chair.

 Starting this we will ask how many committees we could make.

 This would be the same as ( n k ).  We must then factor in the people who may be the chair.

 This is k.

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CONTINUED

 Another way to solve the same thing:  We see that there are n students to be the chair of committees, so we select the chair of each subcommittee out of n students.  There are now n-1 students remaining.

 It follows that there are 2 n-1 ways of filling the remaining committees.

 We now have n*2 n-1 ways of choosing different ways of choosing committees.

 If we look at the identity we were given, this helps show that the answer we got is correct.

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HANK YOU TO OUR MENTOR

 Greg Warrington!

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EFERENCES

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LIDE

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 Proofs That Really Count; Benjamin/Quinn  Introduction to Discrete Mathematics, Burgmeier/Kost

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OMEWORK

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ROBLEMS

 We have a baseball team made up of 6 players.

 A. How many different 3 person batting orders can we have? (order matters, Permutation)  B. How many different 2 person groups of team captains can we have? (order doesn’t matter, Combination)