Math 4408 Lecture1

Two Counting Principles:

Addition Principle: Trivial

Multiplication Principle: If a procedure can be broken into m stages and if there are r outcomes in the first stage,s outcomes in the second stage independent of the choice in the first stage, t outcomes in the third stage again independent of the previous choices then the total procedure has rst... composite outcomes.

Examples:

  1. # Students in two classes (don't neglect overlap - addition must be of disjoint sets)
  2. Tossing Dice (no doubles - 2 ways to solve)
  3. Menu (3 columns - 2 choices)
  4. # of subsets of a set

P(n,r) = n(n-1)(n-2)...(n-r+1)

P(n,n) = n!

Some problems:

1. How many ways are there to pick 2 successive cards from a standard deck of 52 such that:
  1. The first card is an Ace and the second is not a Queen?
  2. The first is a spade and the second is not a Queen?
2. There are 15 different apples and 10 different pears. How many ways are there for Jack to pick an apple or a pear and then for Jill to pick an apple and a pear?

3. There are 50 cards numbered 1 to 50. Two different cards are choosen at random. How many ways can this be done? In how many of these picks will one card be twice the other? [25.49, 25]

4. Use the prime decomposition to show that every square integer has an odd number of divisors (including 1 and itself), and every non-square integer has an even number of divisors. [ p1a1p2a2... pkak has (a1 + 1)(a2 + 1) ...(ak + 1) factors ]

5. How many integers greater than 6600 have distinct digits, not including the digits 7,8,or 9? [ 6.6.5.4.3 + 6.6.5.4.3.2 + 6.6.5.4.3.2.1 = 720.15 = 10,800 ]

C(n,r) = n!/r!(n-r)!

Arrangements (lists - orderings)

In general if n1 selections for first place, n2 for second, ... nr for the r-th place (each selection independent of the preceding) then the number of lists is = n1 n2 ...nr .

If selections are taken from an n-set and repeats (replacement) allowed then the number of lists is = n n n ... n = nr .

If no repeats are allowed then the number of lists is given by
P(n,r) = n(n-1)(n-2)...(n-r+1)

In the special case of complete lists (permuations) we have
P(n,n) = n!

Combinations (selections without order; number of r-sets in an n-set) (r-combinations)
P(n,r) = C(n,r) x P(r,r)

C(n,r) = n!/r!(n-r)!

Also, C(n,r) = C(n, n-r)

and C(n,r) = C(n-1,r-1) + C(n-1,r).

Binomial Theorem.

C(n,m) C(m,k) = C(n,k) C(n-k, m-k).

Pascal's Triangle:

C(n-1,k-1) + C(n-1,k) = C(n,k).

C(n+1, k) = C(n, k-1) + C(n-1, k) + C(n-1, k-1).

Pascal's Triangle Mod n (Part I)

Pascal's Triangle Mod n (Part II)