Combinatorial proof!

(n+1 choose k) = n choose (k - 1) + (n choose k).

Combinatorial proof:

1) n+1 choose k is the number of ways to choose k elements out of a set
of n+1 elements.

2) Consider a group of n+1 elements.
     Make one element special, 
   Then to choose k elements, we could either choose the special
     element or not.  
   If we don't choose the special element, there are:
     n choose k ways of choosing the k elements
   If we do choose the special element, there are:
     n choose k-1 ways of choosing the "other" k-1 elements.

   These sets are not intersecting (your subset of k includes the
   special element iff you are in the second option), so the total
   number of ways of choosing k elements from n+1 is:
     n choose k +  n choose k-1.

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Number of ways to make 1's and 2's sum to 12:
Number of ways to make 1's and 2's and 3's sum to 12:
Number of ways to make 2's and 3's sum to 12:

(here the order of the numbers matters!!).

Numbers of ways to define a set of numbers that sums to 12
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Among other things, the Catalan numbers describe

* the number of ways a polygon with n+2 sides can be cut into n triangles;
* the number of ways in which parentheses can be placed in a sequence of numbers to be multiplied, two at a time;
* the number of planar binary trees with n+1 leaves; and
* the number of paths of length 2n through an n-by-n grid that do
    not rise above the main diagonal.

http://mathforum.org/advanced/robertd/catalan.html


Cn+1 = sum(from 0..n) CiCn-1

turns out to equal:
(2n choose n) / n+1

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The goal here is to introduce you to several very different ways of
transforming a combinatorial problem.  Finding such a transformation
is often the hardest part of a counting problem (and, in general, may
be the hardest part of many mathematical or algorithmic problems!).

 How many ways can 4 non-negative integers sum to 12 (order
matters!)? == (12 + 4 - 1 choose 4)

0 + 0 + 5 + 7
7 + 0 + 0 + 5
3 + 3 + 3 + 3

Trick is to change the representation,  and turn this question into a
bit string!  How can we represent 4 integers summing to 12 as a bit
string.  First, use 0's to write out 12 in unary.  Then use 1's to
break the bit string up.

 How many ways can 4 positive integers sum to 12 (order matters!)?
== (12 - 1 choose 4)

here, we aren't allowed to have two 1's in a row (because that would
constitute a '0' in the sum.
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Another representational change.

"During a month with 30 days, a baseball team plays at least one game
a day, but no more than 45 games.  Show there must be a period of
consecutive days during which the team plays exactly 14 games".

Why?  (I DON'T KNOW why this problem is important.).

let gi be the number of games played on day i.

let ai be the number of games played up to day i.
  so ai = sum g1..i. (bad notation!).

Note that all the ai are different (played at least one
                                               game per day).

Note that all the a30 = 45

let's make up a new set of numbers B:

B = {a1,a2, ... and:
     a1+14, a2+14 and so on.}

How many elements does B have?    60
What is the min element of B?     1
What is the max element of B?   45+14 = 59

so, two elements of B must be the same.

Can they be two elements ai?  (no) 
Can they be two elements ai+14?  (no) 
so they must be 1 of each kind:
so there must be some ai = aj+14
so 14 games took place between day i and day j.
 
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Not quite counting... Ramsey theory.

6 friends.  How many different ways can those 6 people be friends or
not?

Prove that all ways that they can be friends, there are either 3
mutual friends or 3 mutual non-friends.

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Birthday probability:

number of ways that we can all have birthdays:        365n.  why?

number of ways that we can have different birthdays:  P(365,n).

So the probability that we all have diff birthdays is:
P(365,n) /  365n.

So the probability that we don't all have diff. birthdays is:

1-(P(365,n) /  365n).

Turns out that for n = 23, this is about 50%.