A license plate consists of a combination of 6 digits or letters. All numbers (09) 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.
Combinatorics Question
This topic has expert replies

 Senior  Next Rank: 100 Posts
 Posts: 47
 Joined: 08 Sep 2009
 Thanked: 2 times
 GMAT Score:540

 Master  Next Rank: 500 Posts
 Posts: 399
 Joined: 15 Apr 2009
 Location: india
 Thanked: 39 times
IMO A
for 1st letter 36 choice
same for rest 5 has 36 choices
hence 36*36*36*36*36*36
OA Pls
for 1st letter 36 choice
same for rest 5 has 36 choices
hence 36*36*36*36*36*36
OA Pls
It does not matter how many times you get knocked down , but how many times you get up

 Senior  Next Rank: 100 Posts
 Posts: 47
 Joined: 08 Sep 2009
 Thanked: 2 times
 GMAT Score:540
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?
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?

 Master  Next Rank: 500 Posts
 Posts: 399
 Joined: 15 Apr 2009
 Location: india
 Thanked: 39 times
000000,aaaaaa,bbbbbb all are uniquePerminology wrote: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?
your method doesn't include these cases
these cases are excluded if and only if its mention that repetetion is not allowed..
It does not matter how many times you get knocked down , but how many times you get up

 Senior  Next Rank: 100 Posts
 Posts: 47
 Joined: 08 Sep 2009
 Thanked: 2 times
 GMAT Score:540
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 09). 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)
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 09). 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)

 Junior  Next Rank: 30 Posts
 Posts: 17
 Joined: 12 May 2008
 Location: India
IMO (d)anu8 wrote: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 09). 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)
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?