Combinatorics CSC 172 SPRING 2002 LECTURE 11

Assignments With Replacement  Example: Passwords  Are there more strings of length 5 built from three symbols or strings of length 3 built from 5 symbols?

In general We are given n “items”, to each we must assign one of k “values” Each value may be used any number of times Let W(n,k) be the number of ways How many different ways may we assign values to the items? “Different ways” means that one or more of the items get different values.

Inductive Definition Basis: W(1,k) == k Induction: If we have n+1 items, we assign the first one in one of k ways and the remaining n in W(n,k) ways

Recurrence W(1,k) = k W(n+1,k) = k * W(n,k) T(1) = k T(n) = k * T(n-1) Easy expansion T(n) = k n

Example: Are there more strings of length 5 built from three symbols or strings of length 3 built from 5 symbols? Length 5, from 3 “values” = {0,1,2} “items” are the 5 positions N = 3 5 = 243 Length 3, from 5 “values” = {0,1,2,3,4} “items” are the 3 positions N = 5 3 = 125

Permutations Example: “SCRABBLE” - start with 7 letters (tiles) - how many different ways can you arrange them - we don’t care about the word’s legality

Scrabble We can pick the first letter to be any of the 7 tiles For each possible 1 st letter, there are 6 choices of second letters Or, 7*6 = 42 possible two letter prefixes Similarly, for each of the 42, there are 5 choices of the third letter. 42 * 5 = 210, and so on Total choices = 7*6*5*…*1 = 7! = 5040 In general, there are n! permutations of n items.

Ordered Selections Suppose we want to begin Scrabble with a 4 letter word? How many ways might we form the word from our 7 distinct tiles? For each possible 1 st letter, there are 6 choices of second letters 7*6 = 42 possible two letter prefixes For each of the 42, there are 5 choices of the third letter. 42 * 5 = 210 For each of the 120, there are 4 choices of the third letter. 210 * 4 = 840

In general  (n,m), the nmber of ways to pick a sequence of m things out of n == n*(n-1)*(n-2)*…*(n-m+1) == n!/(n-m)!

Combinations Suppose we give up trying to make a word and want to throw 4 of our 7 tiles back in the pile? How many different ways can we get rid of 4 tiles? Ordered selection 4!/(7-4!) = 840 However, we don’t care about order. So, how many ways are there to order 4 items? 4! = /24==35 == 7!/((7-4)!4!) = 35

In general “n choose m”

Recursive Definition for n choose m We want to choose m things out of n, we can either take or reject the first item. If we take the first, then we can take the rest by choosing m-1 of the remaining n-1 We can do this in (n-1) choose (m-1) ways OTOH, if we reject the first item, then we can get the rest by choosing m of the remaining n-1 We can do this in (n-1) choose m ways

Inductive Definition Basis: for all n there is only one way to choose all or none of the elements Induction: for 0 < m < n

Proving Inductive Definition = Direct Definition What is the induction parameter? Zero for the basis case decreases in the inductive step Complete induction on m(n-m) Prove: c(n,m) = n!/((n-m)!m!)

Basis If m(n-m) == 0, then either m == 0 or m == n If m == 0, then n!/(n-m)!m! == n!/n! = 1 = c(n,0) If m == n, then n!/(n-m)!m! == n!/n! = 1 = c(n,n)

Induction By definition c(n,m) = c(n-1,m-1) + c(n-1,m) Assume, c(n-1,m-1) = (n-1)!/((n-m)!(m-1)!) c(n-1,m) = (n-1)!/((n-m-1)!m!) Add the left sides == c(n,m), by definition Add the right sides == n!/(n-m)!m!, clearly

Orders With Some Equivalent Items In real life, we play Scrabble with duplicate letters Suppose you draw {S,T,A,A,E,E,E} at star, how many 7-letter “words” can you make. Similar to permutations, but now, some are indistinguishable, because of duplicates Trick: we can mark the letters to make them distinguishable S,T,A 1,A 2,E 1,E 2,E 3 Then we get 7!=5040 ways

How much “sameness” But some order are the same E 3 TA 1 E 1 SA 2 E 2 == E 3 TA 2 E 1 SA 1 E 2 The two As can be ordered in 2! = 2 ways The three Es can be ordered in 3! = 6 ways So, the number of different words is 7!/2!3! = 540/(2*6) = 420

In general The orders of n items with groups i 1,i 2,…,i k equivalent items is n!/(i 1 !i 2 !..i k !)

Items into Bins Suppose we throw 7 dice (6 sided). How many outcomes are there? Place each of 7 items into one of 6 bins The tokes are the dice The bins are then number of dice Putting the second token into bin 3 means that the second die shows 3

Trick Imagine 5 markers, denoted “*” that represent separation between bins, and 7 tokens “T” that represent dice The string “*TT**TTT*T*T” corresponds to no 1s, two 2s, no 3s, three 4s, a 5 and a 6 How many such strings? “orders with identical items” 12 items, 5 of type “*”, 7 of type “T” 12!/5!7! = 792

In general The number of ways to assign n items to m bins The number of orders of n-1 markers and n tokens We are picking n out of the possible n+m-1 positions for the tokens

Several kinds of items into Bins Order in bin can be either important (queues) or not

Several kinds of items, order within bin unimportant If we have items of k colors, with i j items of the jth color, then the number of distinguishable assignments into m bins is: Example: 4 red dice, 3 blue dice, 2 green dice 6 bins

Order important If there are m bins into which ij items are placed and order within the bin matters Imagine m-1 markers separating the bins and i j tokens of the jth type By “orders with identical items”