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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
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2+0 Answers

 #1
avatar+79894 
+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

 

 

cool cool cool

CPhill  Nov 27, 2017
 #2
avatar+79894 
+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

 

 

cool cool cool

CPhill  Nov 27, 2017

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