15B.) Solve the following equations algebraically for n *Show working using factorial notation
nP4 = 84 nC2
11.) A teacher makes a multiple-choice quiz with 12 questions. 3 answers are A, 3 answers are B, 2 answers are C and 4 answers are D. How many possible answer keys are possible?
14a.) (n-4)! / 3! (n-2)!
edit: I think the answer is n=9
b.) nC3 + nC2 + nC1
+128407 nP4 = 84 nC2
n! / (n - 4)! = 84 n! / [ (n - 2)! * 2!]
1 / ( n - 4)! = 42 / (n - 2)!
(n - 2)! / ( n - 4)! = 42
(n - 2) (n - 3) = 42
n^2 - 5n + 6 = 42
n^2 - 5n - 36 = 0
(n - 9) ( n + 4) = 0
Setting each factor to 0 and solving for n.......n = 9 or n = -4
Take the positive answer ⇒ n = 9
