combinatorics

[rishianand7]

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

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

[Brent@GMATPrepNow]

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

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

NOTE: I'm assuming that the license plate ABC123 is different from A1B23C, even though they use the same characters.

Take the task of building a license plate and break it into stages.

Stage 1: Select a character for the 1st position.
There are 36 characters to choose from (26 letters + 10 digits)
So, we can accomplish this stage in 36 ways.

Stage 2: Select a character for the 2nd position.
Since there is no restriction that prohibits duplicate characters, we can accomplish this stage in 36 ways.

Stage 3: Select a character for the 3rd position.
Since there is no restriction that prohibits duplicate characters, we can accomplish this stage in 36 ways.

Stage 4: Select a character for the 4th position.
We can accomplish this stage in 36 ways

Stage 5: Select a character for the 5th position.
We can accomplish this stage in 36 ways

Stage 6: Select a character for the 6th position.
We can accomplish this stage in 36 ways

By the Fundamental Counting Principle (FCP) we can complete all 6 stages (and thus build a licensne plate) in (36)(36)(36)(36)(36)(36) ways ([spoiler]= 36^6 ways = A[/spoiler])

Cheers,
Brent

Aside: For more information about the FCP, we have a free video on the subject: https://www.gmatprepnow.com/module/gmat-counting?id=775

[rakeshd347]

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

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

Even if there is some restrictions that you can use one number of digit only once then the answer should be C.
This is my understanding:

for first place 36ways
for second place 35 ways and for 3rd 34, for 4th 33 and so on.

So it will be 36*35*34*33*32*31 which is equal to 36!/30!

[Brent@GMATPrepNow]

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

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

Even if there is some restrictions that you can use one number of digit only once then the answer should be C.
This is my understanding:

for first place 36ways
for second place 35 ways and for 3rd 34, for 4th 33 and so on.

So it will be 36*35*34*33*32*31 which is equal to 36!/30!

You're absolutely right.
IF question stated that we cannot use a character more than once, then the answer would be C.

Cheers,
Brent

[sanjoy18]

Actually this is a permutation problem with repetition

so question becomes .how many ways 6 place can be filled up using 36 items(10 digits+ 26 letters)where repetition is allowed

- - - - - -
36*36*36*36*36*36

hence 36^6..Ans A