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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)

Thanks,
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
problem.

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
expression.

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

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
http://mathforum.org/dr.math/
```
Associated Topics:
College Discrete Math
High School Discrete Mathematics
High School Permutations and Combinations

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