The equation P(n, 3) = 6 C(n, 5) can be solved by manipulating factorial expressions. The correct approach involves cross-multiplying and simplifying the equation to derive n^2 - 7n - 8 = 0. Factoring this quadratic equation yields the solutions n = 8 and n = -1, with n = 8 being the valid solution in the context of permutations and combinations. The discussion highlights the importance of careful notation and the correct application of factorial properties.
PREREQUISITESMathematics students, educators, and anyone interested in combinatorial problem-solving techniques.
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}