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

\(A. 36^6\)

\(B. \frac{36!}{30!6!}\)

\(C. \frac{36!}{30!}\)

\(D. \frac{36!}{6!}\)

\(E. 30!\)

So the plate is like: XXXXXX. Now, each X can take 10 digits + 26 letters = 36 different values (options), thus total 36^6 unique plates are possible.

Answer: A.

Hello Bunuel

I have a doubt here....... First X can not take 10 values because 0 cant be placed at first place.If we place 0 at the first place the number will become five digit and I think it will not be acceptable in this case. How we will address this case? please explain.
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Bunuel
gmatpapa
A license plate consists of a combination of 6 digits or letters. All numbers (0-9) and all 26 letters may be used. How many unique license plates are there?

\(A. 36^6\)

\(B. \frac{36!}{30!6!}\)

\(C. \frac{36!}{30!}\)

\(D. \frac{36!}{6!}\)

\(E. 30!\)

So the plate is like: XXXXXX. Now, each X can take 10 digits + 26 letters = 36 different values (options), thus total 36^6 unique plates are possible.

Answer: A.

Hello Bunuel

I have a doubt here....... First X can not take 10 values because 0 cant be placed at first place.If we place 0 at the first place the number will become five digit and I think it will not be acceptable in this case. How we will address this case? please explain.

First of all, any of the 6 X's can take 10 digits + 26 letters = 36 different values, not only 10. Next, we have a license plate, not a 6-digit number, which means that the first digit can be 0.
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Bunuel
gmatpapa wrote:
A license plate consists of a combination of 6 digits or letters. All numbers (0-9) and all 26 letters may be used. How many unique license plates are there?

A.366

B.36!30!6!

C.36!30!

D.36!6!

E.30!

So the plate is like: XXXXXX. Now, each X can take 10 digits + 26 letters = 36 different values (options), thus total 36^6 unique plates are possible.

Answer: A.

The question stem says "unique combinations" only. So doesn't that mean there shouldn't be repetition of the digits or alphabets? How are we authenticating the uniqueness in this problem? Could you please explain?

Using My approach, answer was:
36!/30!
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Pretz
Bunuel
gmatpapa wrote:
A license plate consists of a combination of 6 digits or letters. All numbers (0-9) and all 26 letters may be used. How many unique license plates are there?

A.366

B.36!30!6!

C.36!30!

D.36!6!

E.30!

So the plate is like: XXXXXX. Now, each X can take 10 digits + 26 letters = 36 different values (options), thus total 36^6 unique plates are possible.

Answer: A.

The question stem says "unique combinations" only. So doesn't that mean there shouldn't be repetition of the digits or alphabets? How are we authenticating the uniqueness in this problem? Could you please explain?

Using My approach, answer was:
36!/30!

Hi Pretz,

When the question talks about unique license plates, it means that any two license plates should not be having the exact same order of characters. Uniqueness here does not mean that the characters within the license plate number should be different.

For example a license plate having the number 112233 would be different from 221133 and both are acceptable license plate numbers as there is no constraints on repetition of characters.

In cases where the characters also have to be unique within the license plate, the question prompt would give you a constraint that the characters can't be repeated.

In this case as each place on the license plate can take 36 values and there are 6 such places, there would be a total of \(36^6\) unique license plates.

Hope it's clear :)

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Hi Pretz,

There might be some 'regional bias' in this question (especially if you've never seen a license plate before), but a license plate is essentially just a unique 'identifier' for a vehicle. Since the prompt does NOT state that there are any restrictions on 'where' a number or letter can be placed, we have to assume that each number and each letter can appear ANYWHERE and can ALSO REPEAT.

So, the first few license plates would be...
000000
000001
000002
000003
Etc.

And the last few license plates would be...
ZZZZZZ
ZZZZZY
ZZZZZX
Etc.

In this way, each of the 6 "spots" on the license plate could hold ANY of the 36 options (the 10 digits and 26 letters) and placing a number or letter in a 'spot' has NO effect on what can be placed in the other 'spots.' This means that we have 36^6 possible license plates.

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Pretz is right. The question says "combinations", and the word "combination" has a precise definition in mathematics. A combination is a selection in which order does not matter.

Clearly, the question doesn't mean to talk about 'combinations', but it's perfectly understandable that someone familiar with the language of combinatorics would find the wording confusing.
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A
each plate letter can be made of 36 digits/letters - 36^6
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gmatpapa
A license plate consists of a combination of 6 digits or letters. All numbers (0-9) and all 26 letters may be used. How many unique license plates are there?

A. 36^6

B. \(\frac{36!}{30!6!}\)

C. \(\frac{36!}{30!}\)

D. \(\frac{36!}{6!}\)

E. \(30!\)

I have a doubt here.
choosing 6 objects from a set of 36 objects is 36C6. isnt this what we're asked to do in this question?
In other words, why cant B be the answer.
Please explain.
Thank you
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gmatpapa
A license plate consists of a combination of 6 digits or letters. All numbers (0-9) and all 26 letters may be used. How many unique license plates are there?

A. 36^6

B. \(\frac{36!}{30!6!}\)

C. \(\frac{36!}{30!}\)

D. \(\frac{36!}{6!}\)

E. \(30!\)

I have a doubt here.
choosing 6 objects from a set of 36 objects is 36C6. isnt this what we're asked to do in this question?
In other words, why cant B be the answer.
Please explain.
Thank you

Hi cKaur,

36C6 does only the selection of 6 DISTINCT objects out of 36 but neither does 36C6 include any Arrangement which question requires Nor does the question mention that all 6 objects making number plates must be distinct i.e. The repetition of Numbers and letters is allowed.

Hence this case requires Arrangement considering the repetition of numbers and letters is allowed

so Six places of Number plate _ _ _ _ _ _ can be filled in 36 x 36 x 36 x 36 x 36 x 36 = \(36^6\) ways


If the question were that all the numbers and letters used in number plates MUST be distinct then the solution would have included selection of 6 out of 36 objects in 36C6 ways and the by Arranging the selected 6 objects further in 6! ways
Hence answer in that case would have been 36C6 x 6! = 36!/30!


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

A. 36^6

B. \(\frac{36!}{30!6!}\)

C. \(\frac{36!}{30!}\)

D. \(\frac{36!}{6!}\)

E. \(30!\)

I have a doubt here.
choosing 6 objects from a set of 36 objects is 36C6. isnt this what we're asked to do in this question?
In other words, why cant B be the answer.
Please explain.
Thank you

Hi cKaur,

36C6 does only the selection of 6 DISTINCT objects out of 36 but neither does 36C6 include any Arrangement which question requires Nor does the question mention that all 6 objects making number plates must be distinct i.e. The repetition of Numbers and letters is allowed.

Hence this case requires Arrangement considering the repetition of numbers and letters is allowed

so Six places of Number plate _ _ _ _ _ _ can be filled in 36 x 36 x 36 x 36 x 36 x 36 = \(36^6\) ways


If the question were that all the numbers and letters used in number plates MUST be distinct then the solution would have included selection of 6 out of 36 objects in 36C6 ways and the by Arranging the selected 6 objects further in 6! ways
Hence answer in that case would have been 36C6 x 6! = 36!/30!


I hope it clears your doubt.

Yes. I got it now. Thank you so much.
This concept will help me in other questions too.! :)
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ckaur
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ckaur

I have a doubt here.
choosing 6 objects from a set of 36 objects is 36C6. isnt this what we're asked to do in this question?
In other words, why cant B be the answer.
Please explain.
Thank you

Hi cKaur,

36C6 does only the selection of 6 DISTINCT objects out of 36 but neither does 36C6 include any Arrangement which question requires Nor does the question mention that all 6 objects making number plates must be distinct i.e. The repetition of Numbers and letters is allowed.

Hence this case requires Arrangement considering the repetition of numbers and letters is allowed

so Six places of Number plate _ _ _ _ _ _ can be filled in 36 x 36 x 36 x 36 x 36 x 36 = \(36^6\) ways


If the question were that all the numbers and letters used in number plates MUST be distinct then the solution would have included selection of 6 out of 36 objects in 36C6 ways and the by Arranging the selected 6 objects further in 6! ways
Hence answer in that case would have been 36C6 x 6! = 36!/30!


I hope it clears your doubt.

Yes. I got it now. Thank you so much.
This concept will help me in other questions too.! :)

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Why is this a hard question? I think this should be sub-600 level.
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Why is this a hard question? I think this should be sub-600 level.

Because This question comes from the topic "Permutation & combination" which might not be difficult for you but it is difficult for many other students.

You can see the posts above and can conclude that some students get confused in such questions as well and there are plenty of such students who don't write their confusion but they are confused.
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I do find the topic of permutation and combination difficult actually. But that is only when there are restrictions implemented in the question. That is why i found this question so straightforward. But I get your point.
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tjerkrintjema - you were completely right. This is definitely not what a 700-level probability/counting question on the real GMAT is like. If you look at the hardest P+C questions on the actual test, they're a lot trickier than this. If this question were properly worded and appeared on the real test, it would probably be in the 500-550 range.
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Nice question. The fundamental counting rules are being tested:-
We have 6 places to fill from 10 digits and 26 letters.
Now, the first place can be selected from 10 digits OR 26 letters. This implies that first position can be filled with 36 ways (OR implies we need to add).
Then since digits and letters can repeat,
36*36*36*36*36*36 = 36^6.
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Hi guys,
Here the question is asking about the the combination of six digits or letters. As the word OR is here so I have added the combination that can be formed using the digits and using the letters. According to me it is 10⁶+26⁶. But this option is not even in the answer choices. What is my fault. Help me, please.
Thanks in advance.
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