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 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 http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/