Counting Subsets of a Set
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Transcript Counting Subsets of a Set
Counting Subsets of a
Set: Combinations
Lecture 28
Section 6.4
Thu, Mar 30, 2006
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
n!
C (n, r )
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.
Counting r-Combinations
Proof (by induction on n).
Base case:
Let n = 0. Then r = 0 and there is only one
0-combination, the null set.
Also, 0!/(0!0!) = 1.
So the statement is true when n = 0.
Counting r-Combinations
General case:
Suppose that the statement is true when n =
k, for some integer k 0.
Consider a set of k + 1 elements.
If r = 0, then there is only one 0-combination,
the null set, and (k + 1)!/(0!(k + 1)!) = 1.
If r = k + 1, then there is only one kcombination, the entire set, and
(k + 1)!/((k + 1)!0!) = 1.
Counting r-Combinations
So let r be any number between 0 and k + 1
(0 < r < k + 1).
Select an arbitrary element a from the set.
For each r-combination of the k + 1
elements, a is either a member or not a
member.
We will count the r-combinations for which
a is a member and then count the rcombinations for which a is not a member.
Counting r-Combinations
r-combinations for which a is not a member:
• The r elements come from the remaining k
elements.
• Therefore, by the inductive hypothesis, there
are k!/(r!(k – r)!) such sets.
r-combinations for which a is a member:
• The other r – 1 elements in the subset come
from the k remaining elements in the set.
• By the inductive hypothesis, there are
k!/((r – 1)!(k – (r – 1))!) such sets.
Counting r-Combinations
Therefore, the number of r-combinations of k
+ 1 elements is
n!
n!
r!(n r )! (r 1)!(n (r 1))!
A little algebra shows that this equals
(n 1)!
r!(( n 1) r )!
Thus, the statement is true when n = k + 1,
etc.
Example: Counting rCombinations
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.
Example: Counting rCombinations
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.
Example: Counting rCombinations
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.
Example: Counting rCombinations
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.
Example: Counting rCombinations
What if I dropped 44 grades!!!???
This must involve unimaginably many
possibilities!
How many would it be?
Lotto South
In Lotto 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.
Lotto South
There are C(49, 6) = 13,983,816 possible
choices.
Match all 6 numbers
There is only 1 winning combination.
Probability of winning is
1/13983816 = 0.00000007151.
Lotto South
Match 5 of 6 numbers
There 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.
Lotto South
Match 4 of 6 numbers
Player chooses 4 winning numbers and 2
losing numbers.
Number of ways is C(6, 4) C(43, 2) =
13545.
Probability is 0.0009686.
Lotto South
Match 3 of 6 numbers
Player chooses 3 winning numbers and 3
losing numbers.
Number of ways is C(6, 3) C(43, 3) =
246820.
Probability is 0.01765.
Lotto South
Match 2 of 6 numbers
Player chooses 2 winning numbers and 4
losing numbers.
Number of ways is C(6, 2) C(43, 4) =
1851150.
Probability is 0.1324.
Lotto South
Match 1 of 6 numbers
Player chooses 1 winning numbers and 5
losing numbers.
Number of ways is C(6, 1) C(43, 5) =
3011652.
Probability is 0.4130.
Lotto South
Match 0 of 6 numbers
Player chooses 6 losing numbers.
Number of ways is C(43, 6) = 2760681.
Probability is 0.4360.
Lotto South
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.
Lotto South
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,
r n r n
k 0 k r k
r
r
Proof: See p. 362, Sec. 6.6, Ex. 18.
Another Lottery
In the previous lottery, the probability of
winning a cash prize is 0.018637545.
Suppose that the prize for matching 2
numbers is… another lottery ticket!
Then what is the probability of winning a
cash prize?
Lotto South
What is the average prize value of a ticket?
Multiply each prize value by its probability
and then add up the products:
$10,000,000 0.00000007151 = 0.7151
$1000 0.00001845 = 0.0185
$75 0.0009686 = 0.0726
$5 0.01765 = 0.0883
$0 0.9814 = 0.0000
Lotto South
The total is $0.8945, or 89.45 cents
(assuming that the big prize is ten million
dollars).
A ticket costs $1.00.
How large must the grand prize be to make
the average value of a ticket more than
$1.00?
Another Lottery
What is the average prize value if matching
2 numbers wins another lottery ticket?
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.
How many for VIRGINIA?
How many for INDIVISIBILITY?