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)!

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- Oct 9th 2008, 05:23 PMSkozFactorial 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 PMskeeter
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 PMchiph588@
(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 PMSkoz
Can you help me with this one

Solve for n.

P(n,1)=6 - Oct 10th 2008, 02:54 PMPlato
You are missing out on some part of the definition.

$\displaystyle \begin{gathered} n(n - 1)! = n! \hfill \\ 10\left( {9!} \right) = 10! \hfill \\ \end{gathered} $

Thus, $\displaystyle (n - 1)!\left[ {n^2 + n} \right] = (n - 1)!\left[ {n\left( {n + 1} \right)} \right] = \left[ {n(n - 1)!} \right](n + 1) = \left[ {n!} \right](n + 1) = (n + 1)!$ - Oct 10th 2008, 05:05 PMmr fantastic
- Oct 10th 2008, 07:46 PMSkoz
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 PMmr fantastic