6.4.1
Section 4: Counting
Subsets of a Set
• In Section 2, we looked at counting events with
or without repetition, but in either instance the
order of the elements mattered.
• Now, we shall relax the order restriction to allow
counting set structures where events are not
distinguished by the order of elements, but by
the mere clustering of elements together.
• This will lead to the last rule, the Division Rule
(not in the text!).
6.4.2
The Division Rule
• Theorem: Suppose a set A has n elements and is
partitioned by the collection {A1, A2, ..., Ap},
where each partition set has m elements. Then:
p = n/m.
• In other words, if a set is partitioned into equalsized partition sets, then the number of partition
sets is the quotient of the size of the set with the
size of any partition set.
• For example, if a set has 100 elements and is
partitioned in 20-element subsets, then there must
be 5 subsets (equivalence classes).
6.4.3
Counting Subsets
• How many 3-element subsets of a 4-element set
are there:
• Let A = {1,2,3,4} then all 3-permutations are:
123, 132, 213, 231, 312, 321 → {1,2,3}
124, 142, 214, 241, 412, 421 → {1,2,4}
134, 143, 314, 341, 413, 431 → {1,3,4}
234, 243, 324, 342, 423, 432 → {2,3,4}.
• Hence # 3-element subsets
= (# 3-permutations) / (# 3-orderings)
= P(4,3) / 3! = 4! / (1!3!) = 4! / 3! = 4.
6.4.4
Combinations
• What we have just counted is a combination. In
this instance, it was a combination of 4 elements
taken 3 at a time.
• We use the Division Rule to negate the order
condition of the permutation counts.
• In general, C(n,k) = P(n,k) / k!
• Equivalently, we use the “choose” notation to get:
n
k
()=
n!
k!(n − k)!
6.4.5
More Counting Subsets
• How many subsets of a 10-element set have 3
elements? How many have 7 elements?
• Solution:
C(10,3) = 10! / (3!7!)
C(10,7) = 10! / (7!3!), the same!
• Note: Counting subsets containing 3 elements is
the same as counting subsets NOT containing the
other 7 elements!
• Theorem: C(n,k) = C(n,n−k).
• How many subsets have at least 8 elements?
• Solution: C(10,8) + C(10,9) + C(10,10)
6.4.6
Counting Binary Strings
• How many 10-bit strings have three 1’s?
• Solution: We model this as requiring us to choose
three of the ten slots to place a 1 then the other
seven remaining slots will get a 0. Thus the
number of 10-bit strings that have three 1’s is
C(10,3) = 10! / (3!7!).
• In general, the number of n-bit binary strings with
density k is C(n,k).
• As before, having k 1’s is the same thing as
having (n − k) 0’s.
Counting Teams
6.4.7
• Suppose a group of 5 men and 7 women want to
pick a 5-person team.
• How many teams can they make with 3 men and 2
women?
• C(5,3)C(7,2) = [5!/(3!2!)][7!/(5!2!)] = 7!/(3!2!2!).
• How many teams have at least 1 man?
• (All − teams w/no man) = C(12,5) − C(7,5)
= [12!/(7!5!)] − [7!/(5!2!)]
• How many teams have at most 1 man?
• (teams w/no man) + (teams w/1 man)
= C(7,5) + C(5,1)C(7,4) = 7!/(5!2!) + [5!/4!][7!/(3!4!)]
Generalized Permutations
6.4.8
• In Section 2, we learned how to count the number
of orderings of the letters of COMPUTER. What
about the number of orderings of the letters of
MISSISSIPPI?
• In this case, we note that not all the letters are
distinct. In particular,
MISSISSIPPI → IIIISSSSPPM,
so although we are still searching for an ordering
structures, there are sub-unorderings present,
induced by the repeated letters, for which we have
to account.
Generalized Permutations Take 1
6.4.9
• Let’s apply the Division Rule to negate the effect
of the unordering portions of the overall order
problem.
• This leaves us with a total count of 11!/4!4!2!.
• Here, the first quotient of 4! “mods” out the effect
of the unordered I’s, the second quotient of 4!
“mods” out the effect of the unordered S’s, and the
last quotient of 2! “mods” out the effect of the
unordered P’s.
Generalized Permutations Take 2
6.4.10
• If we model this problem, purely as a combination,
and not a permutation at all, we can reason the task
as:
1. Choose 4 slots from 11 for the I’s;
2. Choose 4 slots from the remaining 7
for the S’s;
3. Choose 2 slots from the remaining 3
for the P’s;
4. Place the M (only 1 way remaining).
• This yields: C(11,4)C(7,4)C(3,2)C(1,1) =
(11!7!3!)/(7!4!4!3!2!1!) = 11!/(4!4!2!).
6.4.11
Generalized Permutation Theorem
• Theorem: Suppose a collection consists of n
objects of which:
n1 are of type 1, indistinguishable from each other;
n2 are of type 2, indistinguishable from each other;
...
nk are of type k, indistinguishable from each other;
and n1 + n2 + ... + nk = n. Then the number of
distinct permutations of the n objects is:
C(n,n1)C(n−n1,n2)C(n−n1−n2,n3)...C(nk,nk) =
n! / (n1!n2!n3!...nk!).