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)!
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, $\displaystyle (n - 1)!\left[ {n^2 + n} \right] = (n - 1)!\left[ {n\left( {n + 1} \right)} \right] = \left[ {n(n - 1)!} \right](n + 1) = \left[ {n!} \right](n + 1) = (n + 1)!$
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?