## Combinatorics Question

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### Combinatorics Question

by Perminology » Fri Nov 13, 2009 11:44 pm
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.

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by xcusemeplz2009 » Sat Nov 14, 2009 12:01 am
IMO A

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

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by Abdulla » Sat Nov 14, 2009 12:15 am
what is the OA? I am confused
Abdulla

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by Perminology » Sat Nov 14, 2009 1:33 am
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

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by xcusemeplz2009 » Sat Nov 14, 2009 2:17 am
Perminology 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

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..
It does not matter how many times you get knocked down , but how many times you get up

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by Perminology » Sat Nov 14, 2009 2:43 am
Thanks! That made a lot of sense.

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### Another Combination question

by anu8 » Sun Nov 15, 2009 3:11 am
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)

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### Re: Another Combination question

by h_jitendras » Sun Nov 15, 2009 7:03 am
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 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?

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