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.
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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.

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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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.

Final Answer:

GMAT assassins aren't born, they're made,
Rich
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A
each plate letter can be made of 36 digits/letters - 36^6
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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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We have a tradition here on GMAT CLUB.

"Thank you" is extended by clicking the button on KUDOS on the post that helped you actually.
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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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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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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.