Combinatorics Sum and Product Rules, Study notes of Discrete Mathematics

The sum and product rules in combinatorics, which are techniques used to count the number of ways to do something without actually enumerating all of them. It provides examples such as counting the number of possible license plates and the number of ways to tour all 50 state capitals. The document also discusses subtler examples and how to cope with ambiguity in combinatorial problems. It is a useful study material for university students taking courses in mathematics, computer science, or statistics.

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Combinatorics
Problem: How to count without counting.
How do you figure out how many things there are with
a certain property without actually enumerating all of
them.
Sometimes this requires a lot of cleverness and deep math-
ematical insights.
But there are some standard techniques.
That’s what we’ll be studying.
1
Sum and Product Rules
Example 1: In New Hampshire, license plates consisted
of two letters followed by 3 digits. How many possible
license plates are there?
Answer: 26 choices for the first letter, 26 for the second,
10 choices for the first number, the second number, and
the third number:
262×103= 676,000
Example 2: A traveling salesman wants to do a tour of
all 50 state capitals. How many ways can he do this?
Answer: 50 choices for the first place to visit, 49 for the
second, . . . : 50! altogether.
Chapter 4 gives general techniques for solving counting
problems like this. Two of the most important are:
2
The Sum Rule: If there are n(A) ways to do Aand,
distinct from them, n(B) ways to do B, then the number
of ways to do Aor Bis n(A) + n(B).
This rule generalizes: there are n(A) + n(B) + n(C)
ways to do Aor Bor C
In Section 4.8, we’ll see what happens if the ways of
doing Aand Baren’t distinct.
The Product Rule: If there are n(A) ways to do A
and n(B) ways to do B, then the number of ways to do
Aand Bis n(A)×n(B). This is true if the number of
ways of doing Aand Bare independent; the number of
choices for doing Bis the same regardless of which choice
you made for A.
Again, this generalizes. There are n(A)×n(B)×n(C)
ways to do Aand Band C
3
Some Subtler Examples
Example 3: If there are nSenators on a committee, in
how many ways can a subcommittee be formed?
Two approaches:
1. Let N1be the number of subcommittees with 1 sen-
ator (n), N2the number of subcommittees with 2
senator (n(n1)/2), . . .
According to the sum rule:
N=N1+N2+· · · +Nn
It turns out that Nk=n!
k!(nk)! (nchoose k); this
is discussed in Section 4.4
A subtlety: What about N0? Do we allow sub-
committees of size 0? How about size n?
The problem is somewhat ambiguous.
If we allow subcommittees of size 0 and n, then
there are 2nsubcommittees altogether.
This is the same as the number of subsets of the
set of nSenators: there is a 1-1 correspondence
between subsets and subcommittees.
2. Simpler method: Use the product rule!
4
pf3
pf4
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Combinatorics

Problem: How to count without counting.

  • How do you figure out how many things there are with a certain property without actually enumerating all of them.

Sometimes this requires a lot of cleverness and deep math- ematical insights.

But there are some standard techniques.

  • That’s what we’ll be studying.

1

Sum and Product Rules

Example 1: In New Hampshire, license plates consisted of two letters followed by 3 digits. How many possible license plates are there? Answer: 26 choices for the first letter, 26 for the second, 10 choices for the first number, the second number, and the third number: 262 × 103 = 676, 000

Example 2: A traveling salesman wants to do a tour of all 50 state capitals. How many ways can he do this? Answer: 50 choices for the first place to visit, 49 for the second,... : 50! altogether. Chapter 4 gives general techniques for solving counting problems like this. Two of the most important are:

2

The Sum Rule: If there are n(A) ways to do A and, distinct from them, n(B) ways to do B, then the number of ways to do A or B is n(A) + n(B).

  • This rule generalizes: there are n(A) + n(B) + n(C) ways to do A or B or C
  • In Section 4.8, we’ll see what happens if the ways of doing A and B aren’t distinct.

The Product Rule: If there are n(A) ways to do A and n(B) ways to do B, then the number of ways to do A and B is n(A) × n(B). This is true if the number of ways of doing A and B are independent; the number of choices for doing B is the same regardless of which choice you made for A.

  • Again, this generalizes. There are n(A)×n(B)×n(C) ways to do A and B and C

Some Subtler Examples

Example 3: If there are n Senators on a committee, in how many ways can a subcommittee be formed? Two approaches:

  1. Let N 1 be the number of subcommittees with 1 sen- ator (n), N 2 the number of subcommittees with 2 senator (n(n − 1)/2),... According to the sum rule:

N = N 1 + N 2 + · · · + Nn

  • It turns out that Nk = (^) k!(nn−!k)! (n choose k); this is discussed in Section 4.
  • A subtlety: What about N 0? Do we allow sub- committees of size 0? How about size n? ◦ The problem is somewhat ambiguous. If we allow subcommittees of size 0 and n, then there are 2n^ subcommittees altogether. ◦ This is the same as the number of subsets of the set of n Senators: there is a 1-1 correspondence between subsets and subcommittees.
  1. Simpler method: Use the product rule!
  • Each senator is either in the subcommittee or out of it: 2 possibilities for each senator: ◦ 2 × 2 × · · · × 2 = 2n^ choices altogether

General moral: In many combinatorial problems, there’s more than one way to analyze the problem.

5

How many ways can the full committee be split into two sides on an issue? This question is also ambiguous.

  • If we care about which way each Senator voted, then the answer is again 2n: Each subcommittee defines a split + vote (those in the subcommittee vote Yes, those out vote No); and each split + vote defines de- fines a subcommittee.
  • If we don’t care about which way each Senator voted, the answer is 2n/2 = 2n−^1. ◦ This is an instance of the Division Rule.

6

Coping with Ambiguity

If you think a problem is ambiguous:

  1. Explain why
  2. Choose one way of resolving the ambiguity
  3. Solve the problem according to your interpretation
    • Make sure that your interpretation doesn’t render the problem totally trivial

More Examples

Example 4: How many legal configurations are there in Towers of Hanoi with n rings? Answer: The product rule again: Each ring gets to “vote” for which pole it’s on.

  • Once you’ve decided which rings are on each pole, their order is determined.
  • The total number of configurations is 3n Example 5: How many distinguishable ways can the letters of “computer” be arranged? How about “dis- crete”? For computer, it’s 8!:
  • 8 choices for the first letter, for the second,... Is it 8! for discrete? Not quite.
  • There are two e’s Suppose we called them e 1 , e 2 :
  • There are two “versions” of each arrangement, de- pending on which e comes first: discre 1 te 2 is the same as discre 2 te 1.
  • Thus, the right answer is 8!/2!

It’s useful to define 0! = 1.

Why?

  1. Then we can inductively define (n + 1)! = (n + 1)n!, and this definition works even taking 0 as the base case instead of 1.
  2. A better reason: Things work out right for P (n, 0) and C(n, 0)!

How many permutations of n things from n are there?

P (n, n) =

n! (n − n)!

n! 0!

= n!

How many ways are there of choosing n out of n? 0 out of n?    n n

   = n! n!0!

   n 0

   = n! 0!n!

13

More Questions

Q: How many ways are there of choosing k things from { 1 ,... , n} if 1 and 2 can’t both be chosen? (Suppose n, k ≥ 2.) A: First find all the ways of choosing k things from n— C(n, k). Then subtract the number of those ways in which both 1 and 2 are chosen:

  • This amounts to choosing k−2 things from { 3 ,... , n}: C(n − 2 , k − 2). Thus, the answer is C(n, k) − C(n − 2 , k − 2)

Q: What if order matters? A: Have to compute how many ways there are of picking k things, two of which are 1 and 2. P (n, k) − k(k − 1)P (n − 2 , k − 2)

14

Q: How many ways are there to distribute four distinct balls evenly between two distinct boxes (two balls go in each box)?

A: All you need to decide is which balls go in the first box. C(4, 2) = 6

Q: What if the boxes are indistinguishable?

A: C(4, 2)/2 = 3.

Combinatorial Identities

There all lots of identities that you can form using C(n, k). They seem mysterious at first, but there’s usually a good reason for them. Theorem 1: If 0 ≤ k ≤ n, then C(n, k) = C(n, n − k).

Proof:

C(n, k) = n! k!(n − k)!

n! (n − k)!(n − (n − k))!

= C(n, n−k)

Q: Why should choosing k things out of n be the same as choosing n − k things out of n? A: There’s a 1-1 correspondence. For every way of choos- ing k things out of n, look at the things not chosen: that’s a way of choosing n − k things out of n. This is a better way of thinking about Theorem 1 than the combinatorial proof.

Theorem 2: If 0 < k < n then    n k

   =

   n − 1 k

   +

   n − 1 k − 1

  

Proof 1: (Combinatorial) Suppose we want to choose k objects out of { 1 ,... , n}. Either we choose the last one (n) or we don’t.

  1. How many ways are there of choosing k without choos- ing the last one? C(n − 1 , k).
  2. How many ways are there of choosing k including n? This means choosing k − 1 out of { 1 ,... , n − 1 }: C(n − 1 , k − 1).

Proof 2: Algebraic...

Note: If we define C(n, k) = 0 for k > n and k < 0, Theorems 1 and 2 still hold.