# Factorial Notation.

• Oct 9th 2008, 05:23 PM
Skoz
Factorial Notation.
How would I solve this question?

Simplify, neW

n(n-1)! the answer is suppose to be n!

and

(n-1)!(n^2+n) the answer is suppose to be (n+1)!
• Oct 9th 2008, 05:29 PM
skeeter
n(n-1)! = n[(n-1)(n-2)(n-3)...(3)(2)(1)] = n(n-1)(n-2)(n-3)...(3)(2)(1) = ?
• Oct 9th 2008, 05:34 PM
chiph588@
(n-1)!(n^2+n) = (n-1)!n^2+n(n-1)! = n*n!+n!=n!(n+1) = (n+1)!
• Oct 10th 2008, 02:41 PM
Skoz
Can you help me with this one

Solve for n.

P(n,1)=6
• Oct 10th 2008, 02:54 PM
Plato
Quote:

Originally Posted by Skoz
Simplify, neW
n(n-1)! the answer is suppose to be n!
and
(n-1)!(n^2+n) the answer is suppose to be (n+1)!

You are missing out on some part of the definition.
$\begin{gathered} n(n - 1)! = n! \hfill \\ 10\left( {9!} \right) = 10! \hfill \\ \end{gathered}$
Thus, $(n - 1)!\left[ {n^2 + n} \right] = (n - 1)!\left[ {n\left( {n + 1} \right)} \right] = \left[ {n(n - 1)!} \right](n + 1) = \left[ {n!} \right](n + 1) = (n + 1)!$
• Oct 10th 2008, 05:05 PM
mr fantastic
Quote:

Originally Posted by Skoz
Can you help me with this one

Solve for n.

P(n,1)=6

Step 1: $\frac{n!}{(n-1)!} = 6$
• Oct 10th 2008, 07:46 PM
Skoz
Can someone tell me how I can approach this problem because im not sure if its a formula based question, the answer is suppose to be 5/17.

A bag contains 10 red jellybeans and 8 black jellybeans.

What is the probability that 2 jellybeans selected at random are red?
• Oct 10th 2008, 07:52 PM
mr fantastic
Quote:

Originally Posted by Skoz
Can someone tell me how I can approach this problem because im not sure if its a formula based question, the answer is suppose to be 5/17.

A bag contains 10 red jellybeans and 8 black jellybeans.

What is the probability that 2 jellybeans selected at random are red?