Download Counting Techniques:

Counting Techniques:
r-combinations with
repetition allowed,
Binomial theorem
1
Number of iterations of a nested loop
(First Situation)
 Consider the following nested loop:
for i:=1 to n
for j:=1 to i-1
for k:=1 to j-1
[ Statements]
next k
next j
next i
 Question: How many times the statements in the innermost
loop will be executed?
 Solution: Each iteration corresponds to
a triple of integers (i, j, k) where i > j > k .
The set of all this kind of triples corresponds to
all 3-combinations of {1, …, n} .
Thus, the total number of iterations is C(n,3) .
2
Number of iterations of a nested loop
(Second Situation)
 Consider the following nested loop:
for i:=1 to n
for j:=1 to i
for k:=1 to j
[ Statements]
next k
next j
next i
 Question: How many times the statements in the innermost
loop will be executed?
 Solution: Each iteration corresponds to
a triple of integers (i, j, k) where i ≥ j ≥ k .
Examples: (5, 3, 2), (4, 4, 3), (2, 1, 1), (3, 3, 3) .
How to count the number of this kind of triples?
3
Number of iterations of a nested loop
(Second Situation)
Each triple corresponds to a string of crosses and vertical bars.
• Numbers 1, 2, …, n are considered as categories;
• n-1 vertical bars separate the categories;
• 3 crosses indicate
how many items from each category are chosen.
Examples when n=5:
Category
5
4
3
2
1
 |
|  |  |
|  |  |
|
|
Triple
(5, 3, 2)
|
(4, 4, 3)
|  | 
(2, 1, 1)
Number of iterations of a nested loop
(Second Situation)
• Each triple corresponds to
a string of n-1 vertical bars and 3 crosses.
The length of any string is (n-1)+3 = n+2 .
The number of distinct positions
for the 3 crosses in a string is C(n+2, 3) .
Thus, the number of distinct triples is C(n+2, 3) .
Generalizing,
if the number of nested loops is r
then the number of iterations is C(n-1+r, r) .
5
r-combinations
with repetition allowed
• Definition:
r-combinations with repetition allowed
from a set X of n elements
is an unordered selection of r elements
taken from X with repetition allowed.
• The number of r-combinations with repetition
allowed from a set of n elements is C(n-1+r, r) .
6
Which formula to use?
• We discussed four different ways
of choosing r elements from n .
• The summary of formulas
used in the four situations:
Order matters
Order does not
matter
Repetition allowed
nr
C(n-1+r, r)
Repetition not
allowed
P(n,r)
C(n,r)
Useful formulas for special cases
• C(n,n-r) = C(n,r) .
Proof:
n!
n!
C (n, n  r ) 

 C (n, r )
(n  r )!(n  (n  r ))! (n  r )!r!
• C(n,n)=1
• C(n,n-1) = C(n,1) = n
• C(n,n-2) = C(n,2) = n(n-1) / 2
8
Binomial Theorem
Theorem: For any real numbers a and b
and nonnegative integer n,
n
(a  b) n   C (n, k )  a nk  b k
k 0
 a n  C (n,1)  a n1  b1  C (n,2)  a n2  b 2    C (n, n  1)  a1  b n1  b n
Proof: By distributive law,
(a+b)n can be expanded into products of n letters
where each letter is a or b :
( a  b) n  (a  b)( a  b)  (a  b)



n times
a
a

a  a
a

a b  a
a

a ba  b
b

b












n times
n -1 times
n - 2 times
n times
Binomial Theorem
Proof(cont.): For each k=0,1,…, n,
the product an-kbk occurs as a term in the sum
the same number of times
as the number of ways to choose k positions for b’s.
But this number is C(n,k).
Thus, the coefficient of an-kbk is C(n,k) .
■
Examples: (a+b)2 = a2+C(2,1)∙ab+b2 = a2+2∙ab+b2
(a+b)3 = a3+C(3,1)∙a2b+C(3,2)∙ab2+b3
= a3+3∙a2b+3∙ab2+b3
10
Using the Binomial Theorem
• Question: Which number is larger:
(1.2)4,000 or 800 ?
• Solution: By the binomial theorem,
(1.2)4,000 = (1+.2)4,000
= 14000 + C(4000, 1)∙13999∙.2
+ other positive terms
= 1 + 4000∙1∙.2 + other positive terms
= 1 + 800 + other positive terms > 800
11
Using the Binomial Theorem
n
2  (1  1)   C (n, k )  1
n
n
k 0
nk
n
 1   C (n, k )
k
k 0
 C (n,0)  C (n,1)  C (n,2)    C (n, n)
 total number of subsets
of a set of n elements
12