Question

find n if 4P(n,3)=5P(n-1,3)

Answer 1

Is this permutations problem?

What is the definition of 4P(n,3) ?

Answer 2

yes, it is.

Answer 3

i think 4P(n,3) is a set

Answer 4

By definition
\[P(n,r)= \frac{n!}{(n-r)!}\]
So, for P(n,3), we have \(\frac{n!}{(n-3)!}\), which can be rewritten as
\[\frac{n!}{(n-3)!} = \frac{n(n-1)!}{(n-3)!}=\frac{n(n-1)(n-2)!}{(n-3)!}=\frac{n(n-1)(n-2)(n-3)!}{(n-3)!}\]Then we can cancel the one with factorial, and we have \[P(n,3) = n(n-1)(n-2)\]Can you do something similar for P(n-1,3)?

Answer 5

i'm still kind of lost..sorry

Answer 6

Which part do you not understand?

Answer 7

i was wondering what happened to the coefficient of "4P"
and i can't really figure out how you arrived with n(n-1)!

Answer 8

I think 4P(n,3) = 4 * P(n,3)

Recall n! = n * (n-1) * (n-2) * ... * 3 * 2 * 1
So, (n-1)! = (n-1) * (n-2) * (n-3) * ... * 3 * 2 * 1

Combining the two, we have
n! = n * (n-1) * (n-2) * (n-3) * ... * 3 * 2 * 1
= n * (n-1)!

Answer 9

Example: 7! = 7 x 6!
because
7! = 7 x 6 x 5 x 4 x 3 x 2 x 1
6! = 6 x 5 x 4 x 3 x 2 x 1
So 7! = 7 x 6!

Answer 10

Is something still not clear?

Answer 11

i tried solving for (n-1,3), this is what i got
\[\frac{ (n-1)(n-1)! }{ (n-3)! } = \frac{ (n-1)(n-1)(n-2)(n-3)! }{ (n-3)! } =(n-1)^2(n-2)\]
\[P(n-1,3)=(n-1)^2(n-2)\]

Answer 12

Oh! For \((n-1)!\), we'll have
\[(n-1)! \]\[= (n-1) *(n-2)! \]\[= (n-1)*(n-2)*(n-3)! \]\[= (n-1)*(n-2)*(n-3)*(n-4)!\]

Answer 13

\[
P(n,r)= \frac{n!}{(n-r)!} \\
P(n-1,r)= \frac{(n-1)!}{(n-1-r)!} \\
P(n-1,3)= \frac{(n-1)!}{(n-4)!} = ~~?\\

\]

Answer 14

P(n-1,3)=(n-1)(n-2)(n-3) ?

Answer 15

correct.

Answer 16

You need to solve: 4n(n-1)(n-2) = 5(n-1)(n-2)(n-3)

Answer 17

It is a cubic equation and so it will have three values of n that will satisfy the equation.

Answer 18

can i just divide both sides by (n-1) and (n-2)?

Answer 19

Can't do that because we are solving the equation for n and so when n = 1 or n = 2 we will be dividing by zero.

Answer 20

Instead you can say when n = 1, the LHS is zero and so is RHS. So n = 1 is one solution.
Same thing with n = 2.
There is one more n value (other than 1 or 2) that you need to find.
Now you can divide both sides by (n-1) and (n-2) because we are looking for an n value other than 1 or 2.

Answer 21

\[4n^3-12n^2+8n=5n^3-30n^2+40n-30\]

Answer 22

Oh, no need to do that! You have common factors on both sides and the solution can be easily found as stated in my previous reply.

Answer 23

oh,okay. sorry. so, the value of n is 15?

Answer 24

n = 1
n = 2
n = 15

Answer 25

Sorry, I meant those are solutions to the cubic equation.
But you have to check the original equation to see if there are any extraneous solution.

Answer 26

okay. thank you for your help! :D

Answer 27

The original problem doesn't make sense when n = 1 or n = 2 and so they are extraneous solutions. So I think n = 15 is the answer even though Wolfram lists all three answers.

Answer 28

http://www.wolframalpha.com/input/?i=4P%28n%2C3%29%3D5P%28n-1%2C3%29