The discussion revolves around solving the equation involving permutations and combinations, specifically finding the value of n in the equation P(n, 3) = 6 C(n, 5). Participants explore various approaches to manipulate the equation and express their attempts at solving it.
Participants do not reach a consensus on the solution, as the discussion includes corrections and multiple attempts to solve the equation, indicating ongoing uncertainty and refinement of ideas.
There are unresolved aspects regarding the initial error pointed out by participants and the implications of the quadratic equation derived from the manipulation of the original equation.
NotaMathPerson said:Find n if P(n, 3) = 6 C(n, 5).
My attempt
$\frac{n!}{(n-3)!}=6\frac{n!}{(n-5)5!}$
I don't know how to proceed
\begin{array}{ccc}\text{We have:} & \dfrac{n!}{(n-3)!}\;=\; 6\dfrac{n!}{(n-5)5!}\\ \\<br /> \text{Cross-multiply:} & 5!\,n!\,(n-5)! \;=\;6\,n!\,(n-3)! \\ \\<br /> \text{Divide by 6n!:} & 20(n-5)! \;=\;(n-3)! \\ \\<br /> \text{We have:} & 20(n-5)! \;=\;(n-3)(n-4)(n-5)! \\ \\<br /> \text{Divide by }(n-5)! & 20 \;=\;(n-3)(n-4) \\ \\<br /> \text{Simplify:} & n^2 - 7n - 8 \;=\;0 \\ \\<br /> \text{Factor:} & (n+1)(n-8) \;=\;0 \\ \\<br /> \text{Solve:} & n=\cancel{-1}.\;8<br /> \end{array}NotaMathPerson said:Find n if P(n, 3) = 6 C(n, 5).
My attempt
$\frac{n!}{(n-3)!}=6\frac{n!}{(n-5)5!}$
I don't know how to proceed
As http://mathhelpboards.com/members/mrtwhs/ pointed out there is a typo on the first line. It is fixed on the second line.soroban said:\begin{array}{ccc}\text{We have:} & \dfrac{n!}{(n-3)!}\;=\; 6\dfrac{n!}{(n-5)5!}\\ \\<br /> \text{Cross-multiply:} & 5!\,n!\,(n-5)! \;=\;6\,n!\,(n-3)! \\ \\<br /> \text{Divide by 6n!:} & 20(n-5)! \;=\;(n-3)! \\ \\<br /> \text{We have:} & 20(n-5)! \;=\;(n-3)(n-4)(n-5)! \\ \\<br /> \text{Divide by }(n-5)! & 20 \;=\;(n-3)(n-4) \\ \\<br /> \text{Simplify:} & n^2 - 7n - 8 \;=\;0 \\ \\<br /> \text{Factor:} & (n+1)(n-8) \;=\;0 \\ \\<br /> \text{Solve:} & n=\cancel{-1}.\;8<br /> \end{array}