Solving equation involving permutation and combination

[NotaMathPerson]

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

 

[mrtwhs]

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

First, you have an error on the right hand side. After you fix that, try dividing both sides by $$n!$$, Think about what a factorial is.

 

[soroban]

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

$$\begin{array}{ccc}\text{We have:} & \dfrac{n!}{(n-3)!}\;=\; 6\dfrac{n!}{(n-5)5!}\\ \\<br /> \text{Cross-multiply:} & 5!\,n!\,(n-5)! \;=\;6\,n!\,(n-3)! \\ \\<br /> \text{Divide by 6n!:} & 20(n-5)! \;=\;(n-3)! \\ \\<br /> \text{We have:} & 20(n-5)! \;=\;(n-3)(n-4)(n-5)! \\ \\<br /> \text{Divide by }(n-5)! & 20 \;=\;(n-3)(n-4) \\ \\<br /> \text{Simplify:} & n^2 - 7n - 8 \;=\;0 \\ \\<br /> \text{Factor:} & (n+1)(n-8) \;=\;0 \\ \\<br /> \text{Solve:} & n=\cancel{-1}.\;8<br /> \end{array}$$

 

[topsquark]

$$\begin{array}{ccc}\text{We have:} & \dfrac{n!}{(n-3)!}\;=\; 6\dfrac{n!}{(n-5)5!}\\ \\<br /> \text{Cross-multiply:} & 5!\,n!\,(n-5)! \;=\;6\,n!\,(n-3)! \\ \\<br /> \text{Divide by 6n!:} & 20(n-5)! \;=\;(n-3)! \\ \\<br /> \text{We have:} & 20(n-5)! \;=\;(n-3)(n-4)(n-5)! \\ \\<br /> \text{Divide by }(n-5)! & 20 \;=\;(n-3)(n-4) \\ \\<br /> \text{Simplify:} & n^2 - 7n - 8 \;=\;0 \\ \\<br /> \text{Factor:} & (n+1)(n-8) \;=\;0 \\ \\<br /> \text{Solve:} & n=\cancel{-1}.\;8<br /> \end{array}$$

As http://mathhelpboards.com/members/mrtwhs/ pointed out there is a typo on the first line. It is fixed on the second line.

-Dan