#Ariana Yousafzai, 9/30/2020, Assignment 5
#OK to post


#Question 1


P(0) = 0 due to empty set
P(1) = 1 due to 1 possible word
P(2) = 2 because two possible permutations are {1,2} and {2,1}


As seen in the pattern above, P(n) where n = 1 has all permutations starting with 1. P(n) where n = 2 has all permutations with the first element being 2. This continues for all real numbers n, by which the nth subset has all permutations starting with the first one being n. 


If the first element is taken out of the first generated subset, the subset has all permutations of length n-1 from 2 to n. 


Assuming that p(n) = |P(n)|, subsets without the first element included are p(n-1). Thus, n*p(n-1) = p(n). Thus, p(n) = n! 


#Question 2


p(n,a1,a2,a3)=p(n-a1,a1,a2,a3)+p(n-a2,a1,a2,a3)+p(n-a3,a1,a2,a3)


#Question 4


Student who is clever & good looking
Included in line 1: 2
Excluded in line 2: 1
Included in line 3: 0
Excluded in line 4: 0


Counted 2-1 = 1 time


Student who possesses all qualities 
Included in line 1: 4
Excluded in line 2: 6
Included in line 3: 4
Excluded in line 4: 1


Counted 4-6+4-1 = 1 time