Thread: 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)!

n(n-1)! = n[(n-1)(n-2)(n-3)...(3)(2)(1)] = n(n-1)(n-2)(n-3)...(3)(2)(1) = ?

(n-1)!(n^2+n) = (n-1)!n^2+n(n-1)! = n*n!+n!=n!(n+1) = (n+1)!

Can you help me with this one

Solve for n.

P(n,1)=6

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.

Thus,

Originally Posted by Skoz

Can you help me with this one

Solve for n.

P(n,1)=6

Step 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?

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?

Questions unrelated to a thread need to be asked in a new thread. Otherwise threads get too confusing and messy.

For this question I'd suggest drawing a tree diagram. Then it's esay to see that Pr(RR) = (10/18) (9/17) = 10/34 = 5/17.