Combinatorics Question

[Perminology]

A license plate consists of a combination of 6 digits or letters. All numbers (0-9) may be used and all 26 letters may be used. How many unique license plates are there?

A) 36^6
B) 36!/30!*6!
C) 36!/30!
D) 36!/6!
E) 30!

I'd like to see other peoples take on this question.

[xcusemeplz2009]

IMO A

for 1st letter 36 choice
same for rest 5 has 36 choices

hence 36*36*36*36*36*36

OA Pls

[Abdulla]

what is the OA? I am confused

[Perminology]

Sorry for the late reply. You are correct the OA is A. I was checkin to see what other people thought here because I thought "unique" license plates meant that none of the letters or numbers could repeat. If that was the case then 36^6 wouldn't be the correct answer, but neither would anything else listed. Originally I thought the solution would look something like this

10 numbers and 26 letters
9 numbers and 25 letters
8 numbers and 24 letters
7 numbers and 23 letters
6 numbers and 22 letters
5 numbers and 21 letters
Therefore: 36*34*32*30*28*26

What do you think about this? Poorly worded question or am I being to anal about the verbiage?

[xcusemeplz2009]

Sorry for the late reply. You are correct the OA is A. I was checkin to see what other people thought here because I thought "unique" license plates meant that none of the letters or numbers could repeat. If that was the case then 36^6 wouldn't be the correct answer, but neither would anything else listed. Originally I thought the solution would look something like this

10 numbers and 26 letters
9 numbers and 25 letters
8 numbers and 24 letters
7 numbers and 23 letters
6 numbers and 22 letters
5 numbers and 21 letters
Therefore: 36*34*32*30*28*26

What do you think about this? Poorly worded question or am I being to anal about the verbiage?

000000,aaaaaa,bbbbbb all are unique
your method doesn't include these cases
these cases are excluded if and only if its mention that repetetion is not allowed..

[Perminology]

Thanks! That made a lot of sense.

[anu8]

Hi,

this problems from combinations is a little confusing...can anyone help me?


Every employee in a company X can have a unique 6 digit password, with two begining alphabets (from A to Z) and four numericals (from 0-9). If alphabets/numericals can be repeated, what are the number of passwords that can be created?

a) (2^26)+(4^10)
b) (2^26)*(4^10)
c) (26^2)+(10^4)
d) (26^2)*(10^4)
e) (26^4)*(10^2)

[h_jitendras]

Hi,

this problems from combinations is a little confusing...can anyone help me?


Every employee in a company X can have a unique 6 digit password, with two begining alphabets (from A to Z) and four numericals (from 0-9). If alphabets/numericals can be repeated, what are the number of passwords that can be created?

a) (2^26)+(4^10)
b) (2^26)*(4^10)
c) (26^2)+(10^4)
d) (26^2)*(10^4)
e) (26^4)*(10^2)

IMO (d)
The 1st alphabet can be chosen in 26 ways and since repetition is allowed, even the 2nd alphabet can be chosen again in 26 ways. Following a similar logic for selecting the 4 numbers in the password we can see that each of the 4 numbers can be chosen in 10 ways.

The total no. of possible passwords are therefore (26^2)*(10^4) i.e. option D.

What is the OA?