PYTHON Problem using a recursive method

Pascal's Triangle

Pascal's triangle is a triangular arrangement of numbers where the top row contains the single number 1, and the values in each following (centered) row are the sum of the value(s) in the row above. The following first five rows of Pascal's triangle should help demonstrate the idea:

 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 

By convention, the rows and columns of Pascal's triangle are numbered starting from 0 note that the 0th column of every row contains the value 1. To aid in the computation of edge cases (columns in rows that do not have two values above them), it is also convenient to imagine that there are columns in row 0 extending off in both directions that contain 0s. I.e., we might envision the first row of Pascal's triangle as follows:

... 0 0 0 1 0 0 0 ... 1 1 1 2 1 1 3 3 1 1 4 6 4 1 

Wolfram Mathworld has a good writeup on the properties and provenance of Pascal's Triangle.

Complete the following function, which returns the value to be found in a given row and column of Pascal's triangle.

In [ ]:

def pascal(row, column):

 # YOUR CODE HERE

 raise NotImplementedError()

In [ ]:

# generate the first 10 rows of Pascal's Triangle

for row in range(10):

 print('{: ^45}'.format(' '.join(str(pascal(row, col)) for col in range(row+1))))

In [ ]:

# (5 points)

?

from unittest import TestCase

?

tc = TestCase()

?

tc.assertEqual(pascal(0, 0), 1)

tc.assertEqual(pascal(1, 0), 1)

tc.assertEqual(pascal(2, 1), 2)

tc.assertEqual(pascal(5, 1), 5)

tc.assertEqual(pascal(5, 2), 10)

tc.assertEqual(pascal(10, 5), 252)