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?
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?