Lecture 5
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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

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Lecture 5

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Lecture 5

Counting 4.3,4.4

4.3 Permutations


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


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?

answer: 6!

4.3 Combinations


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).


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!


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.


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.



r 


4.4 Binomial Coefficients

We will now study some properties of the binomial coefficients.

First the Binomial theorem:

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


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


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:

alternative proof: we know:

Now set x=1, y=1....


Some other special cases of

set x=1, y=-1

set x=1, y=2






Pascal’s identity:















This leads to Pascal’s Triangle.

From this we see an easy way to generate all coefficients










More Identities:

VanderMonde’s identity:


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










If we set m=n, r=n then we get:



#(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

get the final answer. 

4.4 Exercise 33 p. 334


one possible path.


How many paths are there from (0,0) to (m,n) with right and up moves as the only

allowed moves?

We need exactly, m up moves and n right moves to end in (m,n).

Let “up” be a “1” and right be a “0”. Thus we need to count the total number

of bit-strings with exactly m 1’s and n 0’s. This is equal to C(m+n,n).

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