SOLUTION: The number of ways one can select three cards from a group of n cards (assuming the order of the selection matters), where n>3 P(n)=n^3-3n^2+2n , is given by . For a certain card t

Question 200686: The number of ways one can select three cards from a group of n cards (assuming the order of the selection matters), where n>3 P(n)=n^3-3n^2+2n , is given by . For a certain card trick a magician has determined that there are exactly 60 ways to choose three cards from a given group. How many cards are in the group?
Answer by Edwin McCravy(20060) (Show Source): You can put this solution on YOUR website!
The number of ways one can select three cards from a group of n cards (assuming the order of the selection matters), where , is given by . For a certain card trick a magician has determined that there are exactly 60 ways to choose three cards from a given group. How many cards are in the group?


Substitute  for  and solve for :




Switch sides of the equation:



Get 0 on the right by subtracting 60 from both sides:



The possible rational zeros of P(n) are � the divisors of 
60.

�1, �2, �3, �4, �5, �6, �10, �12, �15, �30, �60

Try 1 using synthetic division:

1 | 1 -3  2 -60
  |    1 -2   0  
   ------------
    1 -2  0 -60

Nope

Try 2 using synthetic division:

2 | 1 -3  2 -60
  |    2 -2   0  
   ------------
    1 -1  0 -60

Nope

Try 3 using synthetic division:

3 | 1 -3  2 -60
  |    3  0   6  
   ------------
    1  0  2 -54

Nope

Try 4 using synthetic division:

4 | 1 -3  2 -60
  |    4  4  24  
   ------------
    1  1  6 -36

Nope

Try 5 using synthetic division:

5 | 1 -3  2 -60
  |    5 10  60  
   ------------
    1  2 12   0

That leaves a 0 remainder, so n=5 is a solution

So we have factored the polynomial equation

   as

 

This gives

 and 



There can be no positive solutions of
 since there are
no sign changes.

So the answer is n=5.

Edwin

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