Combination and Permutation

[kd15]

Solve for n: 3(nC3) = (n-1)P3
I am unsure how you are supposed to do this. I can get the side with Permutation to look like [MATH](n-1)!/((n-1)-3)![/MATH] and I'm unsure how you get the side of Combinations but I believe it is [MATH]3n!/((3n-9)!*9!)[/MATH]I'm not sure if for the combinations if you keep r as 3 or if you change it to 9, I change n to 3n so I figured you'd change the r as well.

After that step though I am confused as to how you are supposed to solve for n. Can anyone help?

 

[Steven G]

3(nC3) = 3*(nC3)

 

[Dr.Peterson]

Solve for n: 3(nC3) = (n-1)P3
I am unsure how you are supposed to do this. I can get the side with Permutation to look like [MATH](n-1)!/((n-1)-3)![/MATH] and I'm unsure how you get the side of Combinations but I believe it is [MATH]3n!/((3n-9)!*9!)[/MATH]I'm not sure if for the combinations if you keep r as 3 or if you change it to 9, I change n to 3n so I figured you'd change the r as well.

After that step though I am confused as to how you are supposed to solve for n. Can anyone help?

The left-hand side is just 3 times [MATH]_nC_3[/MATH], which is [MATH]3\cdot\frac{n!}{(n-3)!3!}[/MATH]. Don't try to distribute the multiplication into the arguments of the binomial coefficient.

Then what you'll want to do is to cancel some common factors in the fractions on each side. For example, the RHS is [MATH]_{n-1}P_3 = \frac{(n-1)!}{(n-4)!} = (n-1)(n-2)(n-3)[/MATH]. Do you see why?