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Pascal’s Triangle: a Picture of N-Choose-K! CSCI 2824, Fall 2012! ! ! Assignments • To read this week: Sections 5.3-5.4 (Ensley/Crawley) • PS 2 due today! • Guest lecture Thursday • We’ll go over the answers to PS1 and PS2 next Tuesday. ! ! ! !! Our example from last time: In a room with 20 people in it, how many distinct handshakes are there? The answer is: 20 choose 2, also written C(20,2) in the textbook, or more often: 20 2 N-Choose-K The formula for C(n,k) is: n! ----------------- (n-k)! k! Note: This is the case where order doesn’t matter, and repeats are not allowed. Other instances of this same variation • How many ways can you choose a committee of 3 people from a set of 12? • How many distinct 5-card poker hands are there? • In a graph of 100 vertices (and many edges), how many distinct ways can you select 8 vertices to look for a “clique” in the graph? A couple of notes about n-choose-k • N-choose-K is the same as N-choose-(N-K) For instance: C(12,3) = C(12,9) • Note that it makes sense (given our conventions) that 0! = 1 A practice problem How many ways are there to get a full house in poker? (Three of one kind of card, two of the other.) A couple of other poker problems… • How many ways are there to get 4 of a kind in a 5-card poker hand? (E.g., QH, QSp, QCl, QD, 7D) • How many ways are there to get a flush? (E.g., 2D, 5D, 7D, 8D, JD) A practice problem with a new wrinkle… Assume that you’re a burglar, and you come up to one of the “standard” keypad alarms, which you know takes a code of four values. You see that only keys 1, 2, and 3 have been smudged (indicating that the 4-number code is something like 1123, or 1232, or something like that). How many possible codes do you have to try to gain entry? Multiple subsets… • Suppose we need to split a group of twelve people into three committees of 4. How many ways are there of doing this? There are C(12, 4) ways of choosing the first committee. Then, once we’ve chosen that first group, from the remaining 8 people, there are C(8,4) ways of splitting into two committees of four. So the total number of ways of forming three committees is: (12!/(8!4!)) * (8!/4!4!) = 12!/(4!4!4!) From poker to bridge… In bridge, the entire deck is dealt to four players, each of whom gets 13 cards. How many possible bridge deals are there? Let’s make a diagram of nchoose-k • We’ll make a table: for the nth row of the table, we’ll list: C(n,0), C(n,1), C(n,2) … C(n,n) 1! 1 1 1 1 1 1 1 ! 1! 2 3 4 5 6 7 1! 3 6 10 15 21 1! 4 10 20 35 1! 5 1! 15 6 1! 35 21 7 1! Pascal’s Triangle 1! 1 1 1 1 1 1 1 1! 2 3 4 5 6 7 1! 3 6 10 15 21 1! 4 10 20 35 1! 5 1! 15 6 1! 35 21 7 1! Some quickie patterns… • The first column (and the final column) are all 1s, because n-choose-0 is 1 and nchoose-n is 1. • The second column is the n-choose-1 column, which is just the natural numbers. • The third column is the n-choose-2 column, which is also called the “triangular numbers” • The fourth column is called “the tetrahedral numbers”: Why is each number inside the triangle (not at the edges) equal to the sum of the two above it? What we want to show is: C(n, k) = C(n-1, k) + C(n-1, k-1) Look at it this way: I want to pick a committee of 4 people from a set of 12. There are C(12, 4) ways of doing this. Suppose there’s one guy in the original group of twelve named “Fred”. Now, I can make my committee either by first choosing to include Fred, or first choosing to exclude him. If I choose to include Fred, I’ll have: C(11, 3) ways of creating the remaining part of the committee. If I choose to exclude Fred, I’ll have: C(11,4) ways of creating the committee. So the total number of possible committees must be the same as the sum of those with Fred and those without Fred. Or, in other words: C(12, 4) = C(11,4) + C(11, 3) What happens if we add all the elements in a row? 1! 1 1 1 1 1 1 1 1! 2 3 4 5 6 7 1! 3 6 10 15 21 1! 4 10 20 35 1! 5 1! 15 6 1! 35 21 7 1 This seems to suggest a pattern… C(n,0) + C(n, 1) + … C(n, n) = 2n Why might this be true? Think of it this way • In a set of n elements, how many possible subsets are there? • One way of thinking about this is to say that for each element we can either choose to include or exclude the element in a given subset. So the total number of subsets must be equal to 2n. Another, equally good way to approach the question… • Or, we could say that the total number of subsets of a set of n elements is just the sum of subsets of all possible sizes. It’s the sum of all subsets of size 0, of size 1, of size 2, etc. C(n,0) + C(n, 1) + C(n, 2)… C(n, n) Why are these values called the “binomial coefficients”? • Start with the binomial (x + y) and raise it to successive powers: (x + y)0 = 1 (x + y)1 = x + y (x + y)2 = x2 + 2xy + y2 (x + y)3 = x3 + 3x2y + 3xy2 + y3 (x + y)4 = x4 + 4x3y + 6x2y2 + 4xy3 + y4 So we end up with the binomial theorem: (x + y)n = Σ0<=k<=n C(n,k) xk y(n-k) Proving the binomial theorem two ways: • The “giving n distinct pieces of candy to x girls and y boys” version • The “multiply lots of (x + y) terms together and see how many times you choose x and how many times you choose y” version Substitute 1 for x, 1 for y: (x + y)n = Σ0<=k<=n C(n,k) xk y(n-k) Now substitute -1 for x, 1 for y: (x + y)n = Σ0<=k<=n C(n,k) xk y(n-k) The Hockey Stick Identity 1! 1 1 1 1 1 1 1 1! 2 3 4 5 6 7 1! 3 6 10 15 21 1! 4 10 20 35 1! 5 1! 15 6 1! 35 21 7 1! ! C(n,2) = C(n-1, 1) + C(n-2, 1) +… C(1, 1) Using the Binomial Theorem to Answer a Sports Gambling Problem • As of today, the Detroit Tigers are beating the New York Yankees in the ALCS, 2 games to 0 (it’s a best-of-seven series). • Suppose that in each given game, the Yankees and Tigers are evenly matched. In other words, for a single game, it’s a 50-50 chance that either will win. • Suppose also that you and a friend each put down $50 before the series. You bet on the Yankees, and your friend bet on the Tigers. You have to end the bet today, however, before the series is over. How much of the $100 pot should you receive? • Let the Tigers and Yankees play out 5 more games (even if the series is “determined” before the 5 games are complete). • The Tigers have to win any two games (or more) out of 5 to win the series. The Yankees have to win 4 or 5 games to win the series. Of the 32 possible sequences of wins and losses, the Tigers win in: C(5,2) + C(5, 3) + (C5, 4) + C(5, 5) of them. The Yankees win in: C(5,0) + C(5, 1) of them. Yankees/Tigers 1! 1 1 1 1 1 1 1 1! 2 3 4 5 6 7 1! 3 6 10 15 21 1! 4 10 20 35 1! 5 1! 15 6 1! 35 21 7 1