!PC -- Sage

Here are some ways we defined n! using just Python:

def fact(n): product = 1 while n>1: product *= n n -= 1 return product

fact(10)

def fact(n): product = 1 for f in range(2, n+1): product *= f return product

fact(10)

def fact(n): if n<2: return 1 return n*fact(n-1)

fact(10)

It's good to understand how these definitions work.

If you understand a little programming, you will be able to use Sage much more effectively.

However, Sage also has stuff like ! built in:

factorial(10)

Same for _nP_r and _nC_r.

Here are some ways we expressed them using pure Python:

def P(n,r): return fact(n)/fact(n-r)

P(10, 3)

def P(n,r): if r == 0: return 1 return n*P(n-1,r-1)

P(10,3)

def C(n,r): return fact(n)/(fact(r)*fact(n-r))

C(10,3)

def C(n,r): if r == 0: return 1 return P(n,r)/fact(r)

C(10,3)

However, Sage already contains a function binomial that returns a binomial coefficient:

binomial(10, 3)

We can easily tell Sage that we'd like to substitute C for binomial:

C = binomial

C(10,3)

And now we can define P in terms of C using the fact that _nP_r = _nC_r \cdot r! :

def P(n,r): return C(n,r)*factorial(r)

P(10,3)

Pascal's Triangle

def row(n): return [C(n,r) for r in [0..n]]

row(10)

for n in range(13): row(n)

Here's another cool way to create Pascal's triangle using pure Python:

row = [1] def expand(row): return [a+b for (a,b) in zip([0]+row, row+[0])]

for n in range(13): row row = expand(row)

Sage also has the functions Permutations and Combinations that are worth exploring:

Permutations('abc')

list(_)

list(Permutations('abc',2))

Combinations('abcde')

list(_)

list(Combinations('abcde', 2))

list(Permutations([1..6], 2))

[[1, 2], [1, 3], [1, 4], [1, 5], [1, 6], [2, 1], [2, 3], [2, 4], [2, 5],
[2, 6], [3, 1], [3, 2], [3, 4], [3, 5], [3, 6], [4, 1], [4, 2], [4, 3],
[4, 5], [4, 6], [5, 1], [5, 2], [5, 3], [5, 4], [5, 6], [6, 1], [6, 2],
[6, 3], [6, 4], [6, 5]]

[[1, 2], [1, 3], [1, 4], [1, 5], [1, 6], [2, 1], [2, 3], [2, 4], [2, 5], [2, 6], [3, 1], [3, 2], [3, 4], [3, 5], [3, 6], [4, 1], [4, 2], [4, 3], [4, 5], [4, 6], [5, 1], [5, 2], [5, 3], [5, 4], [5, 6], [6, 1], [6, 2], [6, 3], [6, 4], [6, 5]]

len(_)

[a+b for (a,b) in list(Permutations([1..6],2))]

[3, 4, 5, 6, 7, 3, 5, 6, 7, 8, 4, 5, 7, 8, 9, 5, 6, 7, 9, 10, 6, 7, 8,
9, 11, 7, 8, 9, 10, 11]

[3, 4, 5, 6, 7, 3, 5, 6, 7, 8, 4, 5, 7, 8, 9, 5, 6, 7, 9, 10, 6, 7, 8, 9, 11, 7, 8, 9, 10, 11]

!PC

533 days ago by mpaul

Pascal's Triangle