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P(2n,3) = 2P(n,4)

Date: 10/08/2002 at 19:01:36
From: Rita
Subject: Permutations equation

I have no idea how to solve the following question:

   P(2n,3) = 2P(n,4)

Rita M.

Date: 10/08/2002 at 22:01:26
From: Doctor Greenie
Subject: Re: Permutations equation

Hello, Rita -

You have at least a couple of ways you can go about solving this 

The most obvious path to try is to evaluate the algebraic expressions 
P(2n,3) and 2P(n,4), set them equal to each other, and solve for n:

  (2n)(2n-1)(2n-2) = 2(n)(n-1)(n-2)(n-3)  (1)

This leads to a cubic equation that fortunately has three real roots, 
and so it is not a great deal of work to factor the expression and 
find the roots (only one of which makes sense for the given problem). 
However, I think there is an easier path to the solution.

It is probably less work to simply make a table of small values of n 
and evaluate the two expressions in equation (1) above and see when 
they are equal. If you leave the expressions in their factored forms, 
rather than performing the indicated multiplications to get a single 
numerical value, finding the value of n that satisfies the equation is 
relatively easy, because for most values of n one or the other of the 
expressions contains a prime factor that is not a factor in the other 

Here is such a table, starting with n=4 (because P(n,4) doesn't have 
any meaning for n less than 4) and going to n=12. The answer is in the 
table somewhere; it only requires a bit of regrouping of the factors 
to see where the two expressions are equal.

   n      P(2n,3)       2P(n,4)
   4   (8)(7)(6)     2(4)(3)(2)(1)
   5   (10)(9)(8)    2(5)(4)(3)(2)
   6   (12)(11)(10)  2(6)(5)(4)(3)
   7   (14)(13)(12)  2(7)(6)(5)(4)
   8   (16)(15)(14)  2(8)(7)(6)(5)
   9   (18)(17)(16)  2(9)(8)(7)(6)
  10   (20)(19)(18)  2(10)(9)(8)(7)
  11   (22)(21)(20)  2(11)(10)(9)(8)
  12   (24)(23)(22)  2(12)(11)(10)(9)

Just looking at this table without performing any multiplications, we 
can rule out n=4, n=6, n=7, n=9, n=10, and n=12 immediately, because 
for those values of n the expression for P(2n,3) contains the prime 
numbers 7, 11, 13, 17, 19, and 23, respectively; and these primes do 
not appear in the corresponding expressions for 2P(n,4).

That leaves only n=5, n=8, and n=11 as possible solutions. You could 
at this point perform the indicated multiplications for those cases to 
find where the two expressions are equal; however, you can eliminate 
2 of the 3 possibilities rather quickly by looking further for prime 
factors that occur in one expression and not the other for a 
particular value of n. Then you can either assume that the remaining 
case is the solution to the problem (since I told you the solution was 
in the table somewhere), or you can go ahead and perform the 
multiplication for the remaining case to verify that the numbers are 
the same.

I hope all this helps.  Please write back if you have any further 
questions about any of this.

I had a good time playing with this and seeking different paths to a 
solution - thanks for sending your question to us.

- Doctor Greenie, The Math Forum 
Associated Topics:
College Discrete Math
High School Discrete Mathematics
High School Permutations and Combinations

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