#### Transcript Lec5Counting

```Lecture 5
Counting 4.3,4.4
4.3 Permutations
r-permutation:
An ordered arrangement of r elements of a set of n distinct elements.
Example: S={1,2,3}:
3,2,1 is a permutation of S.
3,2 is a 2-permutation of S
Note: set is unordered, but permutation is ordered! (no curly brackets).
The number of r-permutations of n objects is:
P(n,r)=n(n-1)(n-2)...(n-r+1)=n! / (n-r)!
First object can be chosen in n ways, second in (n-1) ways etc. until n-r+1.
Use product rule to get the result.
(reminder: n! = n(n-1)(n-2)...1).
4.3 Permutations
Example:
A mailman needs to bring 8 packages to 8 cities. He starts at city 1.
How many ways are there to visit the remaining 7 cities?
Pick second city among 7, 3rd among 6 etc. = 7!
How many permutations of the letters “abcdefgh” contain “abc” as a block.
Rename “abc” to B. Now we have:
how many permutations of Bdefgh are there?
4.3 Combinations
r-combination:
An unordered selection of r elements (or subset of size r) of a set of n elements.
Example: S={1 2 3 4}.
{ 3 2 1}={1 2 3} is a r-combination.
The total number of r-combinations of a set of size n is given by the
binomial coefficient: C(n,r)=n! / r! (n-r)! 0<=r<=n
Note that this is equal to P(n,r) / r!
The reason is that P(n,r) counts the total number of ordered arrangements.
However, we are only interested in unordered “arrangements” here.
For every subset of r elements one can exactly construct r! ordered arrangements,
everyone of which is included in P(n,r), but should be considered the same in C(n,r).
We thus overcounted by a factor r!, which we need to divide out.
4.3 Combinations
Remark: the expression n! / (n-r)! r! is inefficient to compute.
However, we can rewrite as follows:
n! / (n-r)! r! = n (n-1) ... (n-r+1) / r (r-1) ... 1
if r < n-r
= n (n-1) ...(r+1) / (n-r) (n-r-1) ... 1 if r > n-r
This must always be an integer (which is not so clear from the equation).
Example:
C(4,2) = number of ways to select 2 objects among 4. (“4-choose-2”).
S={1 2 3 4}{1 2} {1 3}{1 4}{2 3}{2 4}{3 4}
C(4,2)=6=4 3 2 1 / 2 *2.
4.3 Combinations
From the definition it is easy to see: C(n,r) = n! / r! (n-r)!
C(n,n-r) = n! / (n-r)! r!
C(n,r)=C(n,n-r)
However, it can also be proved using combinatorial arguments:
Let S be a set of n elements, and A be a subset of r elements.
The following questions have the same answer:
How many subsets A and how many subsets (S-A) are there?
The reason is that for every subset of r elements, there is exactly
one subset of n-r elements which is the remainder.
Since A has r elements and S-A has (n-r) the result follows.
Example:
In how many ways can we pick 5 players from 10 candidates C(10,5).
4.3 Combinations
2) How many bit-strings of size 10 contain 4 1’s.
We need to place 4 1’s in 10 slots: C(10,4).
3) We need to form a committee of 7 people, 3 from math and 4 from computer
science to develop a discrete math course. There are 9 math candidates and
11 CS candidates. How many possibilities?
 Two separate problems that need to be combined using the product rule.
C(9,3) possibilities for math AND C(11,4) possibilities for CS:
Total = C(9,3) C(11,4) = 27,720.
4.3 Counting
Note: C(n,r), P(n,r), n! etc have like 2^n the potential to grow
very fast with n.
Behavior of C/P for fixed n=12 and growing r.
C(12,r)
P(12,r)
r
r
4.4 Binomial Coefficients
We will now study some properties of the binomial coefficients.
First the Binomial theorem:
n
( x  y)n   C (n, j ) x n j y j  C (n, 0) x n  C (n,1) x n1 y  ...
j 0
n, j  0, j  n
Explanation: The left hand side is a product of n terms: (x+y)(x+y) ....
If I am going to write out their multiplication, a cross-product x^(n-j) y^j
can appear in many different ways. To get the coefficient in front we need to
count in how many ways precisely.
-Start with the first term: x^n. From each term in the product I have to pick the x-term.
This can be done in exactly one way.
- The next term: x^(n-1) y can be done in n ways, because I have n choices for the
y variable, and given that, the x’s are picked from the remaining terms.
-General, I have C(n,j) ways to pick y’s (x follows) or equivalently C(n,n-j) ways to pick x’s
4.4 Binomial Coefficients
Examples:
1) What is the coefficient of x^12 y^13 in the expansion of (x+y)^25 ?
I need to pick 12 x’s from 25 terms: C(25,12)=C(25,13).
2) What is the coefficient of x^12 y^13 in (2x-3y)^25 ?
First replace a=2x and b=-3y.
The coefficient of a^12 b^13 in (a+b)^25 is C(25,12).
thus it follows that: C(25,12) a^12 b^13 = C(25,12) 2^12 x^12 (-3)^13 y^13
coefficient is thus: C(25,12) 2^12 (-3)^13
4.4
We already saw that the cardinality of the power-set of a set with
n elements has 2^n elements.
The total number of elements must be equal to the total number of
subsets with zeros element (empty set) pus with 1 element, etc.
There are precisely C(n,j) ways to pick a subset of j elements from a set
with n elements, thus is follows that:
n
2   C (n, j )
n
j 0
 n  n j j
( x  y)     x y
j 0  j 
n
alternative proof: we know:
Now set x=1, y=1....
n
4.4
 n  n j j
Some other special cases of ( x  y)     x
y
j 0  j 
n
n
j n
(

1)

 0
j 0
 j
n
j n
n
(2)

 3
j 0
 j
set x=1, y=-1
n
set x=1, y=2
4.4
Pascal’s identity:
T
C (n  1, j )  C (n, j  1)  C (n, j )
j
T
n, j  0, j  n
T
j
=
a
S
C(n+1,j)
j-1
+
a
S
C(n.j)
a
S
C(n,j-1)
4.4
From this we see an easy way to generate all coefficients
recursively.
C(4,1)+
C(4,2)=
C(5,2)
4.4
r
More Identities:
VanderMonde’s identity:
C (m  n, r )   C (m, r  j )C (n, j )
j 0
m, n, r  0 r  m, n
Proof:
Let Sn be a set with n elements and Sm be a set with m different elements.
The total number of subsets of r elements of the union of Sn and Sm is C(m+n,r).
However, this must be the same as picking zero elements from Sn and r from Sm
(C(n,0) C(m,r)) plus one from Sn and (r-1) from Sm (C(n,1) C(m,r-1)) etc. until
r from Sn and zero from Sm (C(n,r *C(m,0)). Summing these up we get the
identity. 
remove r
0
r
1
n
r-1
n
m
n
m
+...
m
=
+
If we set m=n, r=n then we get: C (2n, n) 
n
n
j 0
j 0
2
C
(
n
,
n

j
)
C
(
n
,
j
)

C
(
n
,
j
)


n0
4.4
n
C (n  1, r  1)   C ( j, r )
j r
n=7,r=3.
#(1,1,1,1,0,0,0,0) = C(3,3)+
#(?,?,?,?,1,0,0,0) = C(4,3)+
#(?,?,?,?,?,1,0,0) = C(5,3)+
.
#(?,?,?,?,?,?,?,1) = C(7,3).
Combinatorial proof using bit-strings:
Left: the number of bit-strings of length n+1 with r+1 one’s.
This must be equal to the following:
Start with all one’s s.t. last 1 is at position r+1. There is one way to do that.
Now all bitstrings s.t. last 1 is at r+2. There are C(r+1,r) ways to do that.
Generally: bitstrings s.t. last 1 is at r+k+1. There are C(r+k,r) ways to do that.
Repeat until r+k = n+1 : C(n,r) ways to do that.
Finally, all these possibilities are different, so we must add them to