Thread: Probabilities hw

Hi, so I have this math homework due for oct 31. I actually don't need an answer, I only want someone to tell me if my answers are correct or not. thanks!

Ok, i only did 2 questions so far haha

is this correct?: 12!=12!x11!x10!x9!x8!x7!x6!x5!x4!x3!x2!x1! = 479 001 600

and: (14 3) [14 is on top and 3 on the bottom] = 14!/3!(14-3)! = 14!/3!11! = 14x13x12/6 = 364

Originally Posted by xluux

Hi, so I have this math homework due for oct 31. I actually don't need an answer, I only want someone to tell me if my answers are correct or not. thanks!

Ok, i only did 2 questions so far haha

is this correct?: 12!=12!x11!x10!x9!x8!x7!x6!x5!x4!x3!x2!x1! = 479 001 600

and: (14 3) [14 is on top and 3 on the bottom] = 14!/3!(14-3)! = 14!/3!11! = 14x13x12/6 = 364

12! = 12 x 11 x 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 479 001 600, not what you wrote.

Your second question is correct.

thanks! i actually wrote what you just wrote on my paper...my brain is kinda tired

Eid?

Originally Posted by Prove It

12! = 12 x 11 x 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 479 001 600, not what you wrote.

Your second question is correct.

Is this correct? A⁷₄= 7!/(7-4)! = 7!/3!
Q^o2: Access codes to the school’s network were given to students, teachers and technicians in a local high school: students’ codes are composed of 4 different letters followed by 3 different numbers. Professors’ are formed of 2 letters (different or not) followed by 4 numbers (different or not). Technicians’ are made up of 1 letter and followed by 5 different numbers.

1. How many different access code this high school can make?
2. What is the probability that the symbols composing a professors’ code are all different?
3. We know the letters and numbers composing a student’s code but do not know in which order they are disposed. What is the probability of finding the right combination the first time?

a) I had a little trouble finding the answers to this question…but here’s what I found:
Students: L²⁶₄= 26!/(26-4)!=26!/22!
N¹⁰₃= 10!/(10-3)!=10!/7!
LxN= 26!/22! x 10!/7! = (22x10)!/(7x26)! = 220!/182!
Students have the possibility of 220!/182! Different access codes.

Teachers: L=(²⁶₂)= 26!/2!(26-2)!= 26!/2!24!=26x25/2=325
N= (¹⁰₄)= 10!/4!(10-4)!= 10!/4!6! = 10x9x8x7/24= 210
LxN= 325x210 = 68250
Teachers have the possibility of 68250 different access codes.

Technicians: L= 1/26
N= A¹⁰₅= 10!/(10-5)! =10!/5! = 3628800/120
1/26 x 3628800/120 = 15120/13
Techs have the possibility of 15120/13 different access codes
So the whole high school can make 220!/182! + 68250 + 15120/13 different access codes…and I’m stuck here because I don’t know how to sum all of this up =/ a little help please?

b) L²⁶₂= 26!/(26-2)! = 26!/24!
N¹⁰₄= 10!/(10-4)! = 10!/6!

• (26x6)!/(24x10)! = 156!/240!

The probability that a teacher’s access code is composed of different numbers and letters is 156!/240! [correct?]

c) I can't find the right formula to find the answer the this one.