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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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Problem: How to count without counting.
Sometimes this requires a lot of cleverness and deep math- ematical insights.
But there are some standard techniques.
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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:
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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).
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.
Example 3: If there are n Senators on a committee, in how many ways can a subcommittee be formed? Two approaches:
N = N 1 + N 2 + · · · + Nn
General moral: In many combinatorial problems, there’s more than one way to analyze the problem.
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How many ways can the full committee be split into two sides on an issue? This question is also ambiguous.
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If you think a problem is ambiguous:
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.
It’s useful to define 0! = 1.
Why?
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!
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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:
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)
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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.
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.
Proof 2: Algebraic...
Note: If we define C(n, k) = 0 for k > n and k < 0, Theorems 1 and 2 still hold.