+0

Last 4 math questions before test tomorrow, help?

+1
138
2

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

Guest Nov 27, 2017
edited by Guest  Nov 27, 2017
Sort:

2+0 Answers

#1
+84238
+1

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

CPhill  Nov 27, 2017
#2
+84238
+1

14b.) nC3 + nC2 + nC1    =

n! / [ ( n - 3)! * 3! ] + n! / [ (n - 2)! * 2] + n

( [ n (n - 1)(n - 2) / 6 ] + [  3 n * (n - 1)] +  6n )   /  6

[ n ]  [ (n - 1)(n - 2) + 3(n - 1) + 6] / 6

[n]  [ n^2 - 3n + 2 + 3n - 3 + 6 ] / 6

[n] [ n^2  + 5] / 6

[ n^3 + 5n ]  / 6

CPhill  Nov 27, 2017

8 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details