.

Is this correct? A

^{7}_{4}= 7!/(7-4)! = 7!/3!

Q

^{o}2: 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.

- How many different access code this high school can make?
- What is the probability that the symbols composing a professors’ code are all different?
- 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?

I had a little trouble finding the answers to this question…but here’s what I found:

Students: L

^{26}_{4}= 26!/(26-4)!=26!/22!

N

^{10}_{3}= 10!/(10-3)!=10!/7!

LxN= 26!/22! + 10!/7! = (22x10)!/(7x26)! = 220!/182!

Students have the possibility of 220!/182! Different access codes.

Teachers: L=(

^{26}_{2})= 26!/2!(26-2)!= 26!/2!24!=26x25/2=325

N= (

^{10}_{4})= 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= A

^{10}_{5}= 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) L

^{26}_{2}= 26!/(26-2)! = 26!/24!

N

^{10}_{4}= 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!

Is anything above correct at all?