Combinatorics

CSC 172

SPRING 2002

LECTURE 11

Assignments With Replacement

Example: PasswordsAre 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 timesLet 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) = kn

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 = 35 = 243

Length 3, from 5 “values” = {0,1,2,3,4}

“items” are the 3 positions

N = 53 = 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 tilesFor each possible 1st letter, there are 6 choices of

second lettersOr, 7*6 = 42 possible two letter prefixesSimilarly, for each of the 42, there are 5 choices of

the third letter. 42 * 5 = 210, and so onTotal 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 1st 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)!

1*...*)1(*)(

1*...*)(*)1(*...*)2(*)1(*),(

mnmn

mnmnnnnmn

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

840/24==35

== 7!/((7-4)!4!) = 35

In general

“n choose m”

!)!(

!

mmn

n

m

n

Recursive Definition for n choose mWe 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

10

n

nn

m

n

m

n

m

n 1

1

1

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)

m

n

m

nmnc

1

1

1),(

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

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!)!1()!1()!(

)!1()!()!1(!)!1()!1(

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mmnnmmnn

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)!()!1()!1()!1(

mmnmn

mnnmmnn

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mmn

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mmn

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Orders With Some Equivalent ItemsIn 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,A1,A2,E1,E2,E3

Then we get 7!=5040 ways

How much “sameness”

But some order are the same

E3TA1E1SA2E2 == E3TA2E1SA1E2

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 i1,i2,…,ik equivalent items is

n!/(i1!i2!..ik!)

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

TrickImagine 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

n

mn

nm

mn 1

!)!1(

)!1(

We are picking n out of the possible n+m-1positions 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 ij items of the jth color, then the number of distinguishable assignments into m bins is:

j

jk

j i

im 11

Example: 4 red dice, 3 blue dice, 2 green dice6 bins

176,1482

7

3

8

4

9

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 ij tokens of the jth type

By “orders with identical items”

jkj im

mn

1)!1(

)1(