Day 10, MAT115R

1. vertices (or points, which we will think of as the people), and
2. edges (which are also called arcs, connecting two points -- which we will think of as friendships).

Now, if each vertex represents a unique individual (we name them), then Pascal's triangle tells us all the different Facebooks possible, of each type. If everyone's friends with everyone else, then the graphs we can get will look like the complete graphs. They're very friendly graphs!

Now how many different Facebook configurations are there for, say

1 individual (zero arcs, so use the zero row!) 1 possible Facebook 2 individuals (one arc, so use the one row) 2 possible Facebooks: 1 with no friendship, 1 with 1 friendship 3 individuals (three arcs, so use the three row) 8 possible Facebooks: 1 with no friendship 3 with one friendship 3 with two friendships 1 with all friendships 4 individuals (six arcs, so use the six row) 64 possible Facebooks: 1 with no friendship 6 with one friendship 15 with two friendships 20 with three friendships 15 with four friendships 6 with five friendship 1 with all friendship

Each choice of a different set of friendships results in a different Facebook. So we begin by asking how many arcs X are possible, and then asking how many ways can we pick Y arcs from X?

The same process works if we want to decide questions like Pascal's about boys and girls, such as

You have four extra tickets to a concert; in how many ways could you choose a set of four friends from your seven favorite friends? (Notice that it's the same number of ways as the number of ways of excluding three friends: the table is symmetric!)

Use the seven row of Pascal's triangle, and go to the "choose 4" entry. The number of things to choose from tells us the row, and the number of things to choose is the entry in that row, and note that we count the entries across a row starting from 0.

Let's use the formula for computing the entry in any row of Pascal's triangle (choose 0, choose 1, ..., choose \(n\))) to check our answer:

\[ C^n_p = \frac{n!}{p!(n-p)!} \] or \[ {n \choose p} \]

(and we say "\(n\) choose \(p\)").

\[ C^7_4 = \frac{7!}{4!(7-4)!} = \frac{7 \cdot 6 \cdot 5 \cdot 4!}{4!3!} = \frac{7 \cdot 6 \cdot 5}{3!} = \frac{7 \cdot 6 \cdot 5}{6} = 35 \]

Remember that that exclamation point (!) means to multiply all the natural numbers from \(n\) down to 1. So, for example, \[ 6! = 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 720 \] We define \(1! = 1\).

Oddly enough, we also define \(0! = 1\).

\[ (a+b)^n \]

So, for example,

\[ (a+b)^0=1 \] \[ (a+b)^1=1a+1b \] \[ (a+b)^2=1a^2+2ab+1b^2 \] \[ (a+b)^3=1a^3+3a^2b+3ab^2+1b^3 \] \[ (a+b)^4=1a^4+4a^3b+6a^2b^2+4ab^3+1b^4 \]

(see your assigned reading).

I hope that you noticed Pascal's triangle peeking out from the right hand sides above!

Do you remember that the Chinese translation of Yang Hui's triangle has the words "to the 4th power" (etc.) in it? These are the powers, of these "binomials".

If it's a fair six-sided die, the probability of rolling a six is 1/6. The probability of rolling a six is \(a=1/6\); then the probability that you don't roll a six is \(b=5/6\).

Let's roll four time:

\[ (a+b)^4=1a^4+4a^3b+6a^2b^2+4ab^3+1b^4 \] Then \[ \left(\frac{1}{6}+\frac{5}{6}\right)^4=1\left(\frac{1}{6}\right)^4+4\left(\frac{1}{6}\right)^3\left(\frac{5}{6}\right)^1+6\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)^2+4\left(\frac{1}{6}\right)^1\left(\frac{5}{6}\right)^3+1\left(\frac{5}{6}\right)^4 \]

Since every denominator has a \(6^4=1296\) in it, we can re-write the probabilities in perhaps a more intuitive way: \[ \left(\frac{1}{6}+\frac{5}{6}\right)^4= \frac{ 1}{1296} + \frac{ 20}{1296} + \frac{150}{1296} + \frac{500}{1296} + \frac{625}{1296} \]

\[ 1 = P(\text{4 sixes} ) + P(\text{3 sixes} ) + P(\text{2 sixes} ) + P(\text{1 sixes} ) + P(\text{0 sixes} ) \]

So only 1/1296 times will you roll a "yahtze" of four sixes.

You'll roll

• 4 sixes 1 time out of 1296,
• 3 sixes 20 times out of 1296,
• 2 sixes 150 times out of 1296,
• 1 six 500 out of 1296 rolls,
• 0 sixes 625 out of 1296 rolls.

For example, suppose that the probability that you have a cold on a given winter day is \(a=1/100\); then the probability that you don't have a cold is \(b=99/100\).

Let's suppose that winter is 90 days long. What is the probability that you have a cold-free winter? We need to compute the 90th row in the table, which has some really big numbers!

\[ \left(\frac{1}{100}+\frac{99}{100}\right)^{90} \]

• Probability of a cold-free winter (0 colds): \[ 1\left(\frac{99}{100}\right)^{90} \approx 0.4047 \]
• Probability of a one-day cold winter (1 colds): \[ 90\left(\frac{1}{100}\right)^1 \left(\frac{99}{100}\right)^{89} \approx 0.3679 \]
• Probability of a two-day cold winter (uh oh: the first two were easy, but now we need the rest of the 90th row of the triangle!:). Thank God for that formula.... \[ 4005\left(\frac{1}{100}\right)^2 \left(\frac{99}{100}\right)^{88} \approx 0.1654 \]

  If you're wondering where that 4005 comes from, it's the triangular entry of the 90th row of Pascal's triangle (third entry from the left, choosing 2): \[ C^{90}_2 = \frac{90!}{2!88!} = \frac{90 \cdot 89}{2} = 4005 \]

  It's also the third entry from the right: remember that the triangle is symmetric.

• Probability of a three-day cold winter: \[ 117480\left(\frac{1}{100}\right)^3 \left(\frac{99}{100}\right)^{87} \approx 0.0490 \]

  The tetrahedral entry of the 90th row of Pascal's triangle: \[ C^{90}_3 = \frac{90!}{3!87!} = \frac{90 \cdot 89 \cdot 88}{3 \cdot 2} = 117480 \]

• etc.

(The great news is that it reduces to the Fraudini trick!) It seems that the Egyptians didn't mind doubling things: that was easy for them; and that's the key to Egyptian multiplication.

• Let's start with a brief review of the history of a western understanding of Egyptian Mathematics.

• At the heart of Egyptian multiplication is the "Fraudini fact", or the "binary factorization":
  Every natural number is either a power of two, or can be expressed as a sum of distinct powers of two in a unique way.

  (I underline "distinct" because you cannot repeat powers: otherwise you could write, for example,

  6=2+2+2 rather than 6=4+2 (which is the unique binary factorization).

• Again this is parallel to prime factorization (and another factorization which is to come!):
  Every natural number (other than 1) is either prime, or can be expressed as a product of prime numbers in a unique way.

• We'll start with a simple example: multiply 23*42.

  1	42
  2
  4
  8
  16
  32	Too big!

  Now add up those rows marked with an asterix (*), and you'll get the answer (966).

• Let's show that we can do a multiplication in either order, by checking the product 42*23. (What is "too big" in this case?)

  I generally double the larger of the two numbers: as a general rule that makes the table shorter.

• Let's try a longer multiplication. Consider, for example, 321*112:

  1	321
  2
  4
  8
  16
  32
  64
  128	Too big!

  Now add up those rows marked with an asterix (*), and you'll get the answer (35952).

• Examples: Try these:
  1. 43*16
  2. 21*79

Watch Dave Brubeck's leg, counting out the rhythm....