1. Counting Techniques: r-combinations with
Counting Techniques: r-combinations with repetition allowed, Binomial theorem

2. 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) .

3. 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?

4. 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
Triple
 | |  |  |
(5, 3, 2)
|  |  | |
(4, 4, 3)
| | |  | 
(2, 1, 1)

5. 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) .

6. 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) .

7. Repetition not allowed
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
Repetition not allowed nr
P(n,r)
C(n-1+r, r)
C(n,r)

8. Useful formulas for special cases
C(n,n-r) = C(n,r) .
Proof:
C(n,n)=1
C(n,n-1) = C(n,1) = n
C(n,n-2) = C(n,2) = n(n-1) / 2

9. Binomial Theorem Theorem: For any real numbers a and b
and nonnegative integer n,
Proof: By distributive law,
(a+b)n can be expanded into products of n letters where each letter is a or b :

10. 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

11. 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
= C(4000, 1)∙13999∙.2
+ other positive terms
= ∙1∙.2 + other positive terms
= other positive terms > 800

12. Using the Binomial Theorem