9.5 Counting Subsets of a Set:
Combinations
1
r-Combination
An r-combination from a set of n-elements is
an r-subset of an n-set.
Any of the symbols C(n, r ), n Cr , C n, r ,
is
used to count the number of r-subsets from
an n-set
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r-Combinations
Consider the set A={a, b, c, d}. Calculate the
number of r-combinations below by showing the
subsets counted in each case
æ 4ö
æ 4ö
ç ÷ = ç ÷ =
è0ø
è1 ø
æ 4ö
æ 4ö
ç ÷ = ç ÷ =
è2ø
è3 ø
æ 4ö
ç ÷=
è 4ø
3
Counting r-Combinations
Take A={a, b, c, d}
Consider the 3-combination {a, b, c}. Notice that {a, b, c}={a ,c,
b}={b, c, a} etc.
How many 3-permutation are there with the elements from the
r-combination
{a, b, c}?
3!C(4, 3) = P(4, 3)
4!
(4 - 3)!
4!
C(4, 3) =
3!(4 - 3)!
3!C(4, 3) =
4
Formula for C(n , r)
The number of r-combinations of an n-set is
given by
r!C(n, r) = P(n, r)
n!
(n - r)!
n!
C(n, r) =
0 £ r £ n
r!(n - r)!
r!C(n, r) =
5
Examples
1. How many distinct five-person teams can be chosen to
work on a special project from a group of 12 people?
Solution:
The number of distinct five-person teams is the same as
the number of subsets of size 5 (or 5-combinations) that
can be chosen from the set of twelve. This number is
.
Thus there are 792 distinct five-person teams.
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Examples
2. Suppose the group of twelve consists of five men and
seven women.
a. How many five-person teams can be chosen that
consist of three men and two women?
b. How many five-person teams contain at least one
man? (two ways)
c. How many five-person teams contain at most one
man?
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Solution
a. To answer this question, think of forming a team as a
two-step process:
Step 1: Choose the men.
Step 2: Choose the women.
There are
ways to choose the three men out of the five
and
ways to choose the two women out of the seven.
Hence, by the product rule,
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b. How many five-person teams contain at least
one man? (two ways)
This question can be answered either by the
addition rule or by the difference rule.
(Difference Rule) Observe that the set of fiveperson teams containing at least one man
equals the set difference between the set
of all five-person teams and the set of fiveperson teams that do not contain any men.
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10
(addition rule) observe that the set of teams
containing at least one man can be
partitioned as shown
Teams with At Least One Man
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The number of teams in each subset of the
partition is calculated using the method
illustrated in part (a). There are
12
Hence, by the addition rule,
13
c. The set of teams containing at most one
man can be partitioned into the set that does
not contain any men and the set that
contains exactly one man.
Hence, by the addition rule,
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Number of Bit String with Fixed
Number of 1’s
How many 10-bit strings are there with
exactly 3 ones?
Label the entries of the bit string from 1-10.
___ ____ ____ _____ ………….._____ _____
1
2
3
4
9 10
To place 3 ones simply choose three of the entries to place them. For
instance {1,7,10} indicates that 1’s will be placed on those entries; {5, 2, 9}
indicates that 1’s will be placed on entries 5, 2, 9. The other entries are
zeroes.
There are a total of
æ 10 ö
ç
÷
è 3 ø
10-bit strings with 3 ones
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Permutations of a Set With Repeated
Elements
Consider different ordering of the letters in
the word MISSISSIPPI
How many distinguishable orderings are there?
IIMSSPISSIP, ISSSPMIIPIS, PIMISSSSIIP, and so on
Approach 1: Strings with all entries different. Account for repetitions (labels)
Approach 2: Strings and select entries to put each of the different letters in the word
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APPROACH 1
Consider all the entries different and then take
care of repetitions
MI1S1S2 I 2 S3S4 I3P1P2 I 4
There are a total of 11 characters, so 11! different words that
can be made
For each word any reordering of the I’s, S’s, P’s will produce the same
word (disregarding the sub-index). There are 4!4! 2! repeated words.
Total of different words with the letters in MISSISSIPPI is
11!
1!2!4!4!
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APPROACH 2
Consider a 11-word where the entries are the letters in
MISSISSIPPI
___ ____ ____ _____ ………….._____ _____ _____
1
2
3
4
9
10
11
Choose the number of ways to place 1 M
Then choose the number of ways to place 4 I’s
Then choose the number of ways to place 4 S’s
Then choose the number of ways to place 2 P’s
æ
ç
è
æ
ç
è
æ
ç
è
æ
ç
è
11 ö
÷
1 ø
10 ö
÷
4 ø
6 ö
÷
4 ø
2 ö
÷
2 ø
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By the multiplication principle the total number
of different words with the letters in MISSISSIPPI
is
æ 11 öæ 10 öæ 6 öæ 2 ö 11! 10! 6! 2!
ç
֍
֍
֍
÷=
è 1 øè 4 øè 4 øè 2 ø 1!10! 4!6! 2!2! 2!0!
11! 10! 6! 2!
=
1!10! 4!6! 4!2! 2!0!
11!
=
1!4!4!2!0!
11!
=
1!4!4!2!
These is the number of permutations of 11 objects with repetitions
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Formula
11!
P(11;1, 4, 4, 2) =
1!4!4!2!
Permutation with repetitions of n characters of k
different types 1
2
k
n + n +… + n = n
n!
P(n;n1, n2 ,… , nk ) =
n1 !n2 !… nk !
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Exercise
1. How many passwords of length 8 can be
made with the letters A, B, C, D, E, F?
2. How many of those passwords have three
B’s, two E’s, and three F’s?
3. How many of those passwords are
permutations of the letters A, B, C, D, E, F?
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