Counting Subsets of a Set: Combinations Lecture 28 Section 6.4 Wed, Mar 2, 2005

Counting Subsets of a Set: Combinations

Lecture 28

Section 6.4

Wed, Mar 2, 2005

r-Combinations

An r-combination of a set of n elements is a subset of r of the n elements.

The order of the elements does not matter. The 3-combinations of the set {a, b, c, d, e}

are {a, b, c}, {a, b, d}, {a, b, e}, {a, c, d}, {a, c, e}, {a, d, e}, {b, c, d}, {b, c, e}, {b, d, e}, {c, d, e}.

Counting r-Combinations

Theorem: The number of r-combinations of a set of n elements is

C(n, r) = n!/[r!(n – r)!]. Examples:

C(4, 2) = (4 3)/(2 1) = 6.C(10, 3) = (10 9 8)/(3 2 1) = 120.C(1000, 2) = (1000 999)/(2 1) =

499500.

Some Useful Facts

C(n, 0) = 1 for all n 0. C(n, 1) = n for all n 1. Notice that C(n, r) = C(n, n – r). For example,

C(100, 99) = C(100, 1) = 100/1 = 100. Therefore,

C(n, n) = 1 for all n 0.C(n, n – 1) = n for all n 1.

Another Useful Fact

The TI-83 will calculate C(n, r).Enter n.Select MATH > PRB > nCr.Enter r.Press ENTER.The value of C(n, r) appears.

Example: Counting r-Combinations

In Math 121, I used to collect 48 daily homework assignments.

Some assignments count more than others.

I drop the “lowest” 4 homework grades. Which should be dropped: 0 out of 30 or

15 out of 40? I drop the 4 grades that hurt the student’s

average the most.


How can that be determined? Can a computer program make the

determination by brute force (exhaustive checking) within a reasonable amount of time?

There are C(48, 4) = 194,580 possible choices.

A computer can do the math really fast, in say one second.


What if I dropped 6 grades? There would be C(48, 6) = 12,271,512

possible choices. Over 60 times as many. This would require about 60 secs = 1 min.


What if I dropped 12 grades? There would be C(48, 12) = over 69 billion

choices! More than 350,000 as many! This would require almost 350,000 sec =

over 4 days.


What if I dropped 44 grades!!!??? This must involve unimaginably many

possibilities! How many would it be?

Example: Lotto South

In Lotto South (Loot the South?), a player chooses 6 numbers from 1 to 49.

Then the state chooses at random 6 numbers from 1 to 49.

The player wins according to how many of his numbers match the ones the state chooses.

See the Lotto South web page.

Example: Virginia Lottery

There are C(49, 6) = 13,983,816 possible choices.

Match all 6 numbersThere is only 1 winning combination.Probability of winning is

1/13983816 = 0.00000007151

(it ain’t gonna happen)


Match 5 of 6 numbersThere are 6 winning numbers and 43 losing

numbers.Player chooses 5 winning numbers and 1

losing numbers.Number of ways is C(6, 5) C(43, 1) = 258.Probability is 0.00001845 (forget about it).


Match 4 of 6 numbersPlayer chooses 4 winning numbers and 2

losing numbers.Number of ways is C(6, 4) C(43, 2) =

13545.Probability is 0.0009686 (nope).




246820.Probability is 0.01765 (it just might

happen.)




1851150.Probability is 0.1324 (it will happen).




3011652.Probability is 0.4130 (it will happen).


Match 0 of 6 numbersPlayer chooses 6 losing numbers.Number of ways is C(43, 6) = 2760681.Probability is 0.4360 (it will happen).


Note also that the sum of these integers is 13983816.

Note also that the lottery pays out a prize only if the player matches 3 or more numbers.Match 3 – win $5.Match 4 – win $75.Match 5 – win $1000.Match 6 – win millions.


Given that a lottery player wins a prize, what is the probability that he won the $5 prize?

P(he won $5, given that he won)

= P(match 3)/P(match 3, 4, 5, or 6)

= 0.01765/0.01864

= 0.9469.

Example

Theorem (The Vandermonde convolution): For all integers n 0 and for all integers r with 0 r n,

Proof: See p. 362, Sec. 6.6, Ex. 18.

r

k r

n

kr

rn

k

r

0

Permutations of Sets with Repeated Elements

Theorem: Suppose a set contains n1 indistinguishable elements of one type, n2 indistinguishable elements of another type, and so on, through k types, where

n1 + n2 + … + nk = n.

Then the number of (distinguishable) permutations of the n elements is

n!/(n1!n2!…nk!).

Proof of Theorem

Proof: Rather than consider permutations per se,

consider the choices of where to put the different types of element.

There are C(n, n1) choices of where to place the elements of the first type.

Proof of Theorem

Proof: Then there are C(n – n1, n2) choices of

where to place the elements of the second type.

Then there are C(n – n1 – n2, n3) choices of where to place the elements of the third type.

And so on.

Proof, continued

Therefore, the total number of choices, and hence permutations, is

C(n, n1) C(n – n1, n2) C(n – n1 – n2, n3) … C(n – n1 – n2 – … – nk – 1, nk)

= …(some algebra)…

= n!/(n1!n2!…nk!).

Example

How many different numbers can be formed by permuting the digits of the number 444556?

6!/(3!2!1!) = 720/(6 2 1) = 60.

Example

How many permutations are there of the letters in the word MISSISSIPPI?

11!/(4!4!2!1!) = 34650.

Documents

Counting Subsets of a Set: Combinations Lecture 28 Section 6.4 Wed, Mar 2, 2005