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Bunuel
ID codes issued by XYZ high school have 5 characters. First 2 characters of the ID codes are vowels and next 3 characters are digits. If ID code of no two students is same, then what is the maximum number of students in XYZ high school?

A. 12,500
B. 14,400
C. 25,000
D. 144,000
E. 250,000

Basically, this is a problem of counting (repetition allowed).

_ _ _ _ _ = five character code.

1st and 2nd character - vowels - A, E, I, O, and U - 5 possibilities.
3rd, 4th and 5th character digits - 0,1,2,3,4,5,6,7,8,9 - 10 possibilities.

Therefore, total number of characters = 5x5x10x10x10 = 25000.

Answer C.
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In the question it is mentioned, no two IDs should be same

Am confused why repetition is allowed in this case.

For example:

_ _ _ _ _ can take a,a,1,1,1

This can be arranged in only one way and even if it is re arranged, the code would be same.

Can someone throw some light on this...
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In the question it is mentioned, no two IDs should be same

Am confused why repetition is allowed in this case.

For example:

_ _ _ _ _ can take a,a,1,1,1

This can be arranged in only one way and even if it is re arranged, the code would be same.

Can someone throw some light on this...

The question says that ID codes of no two students should be same and nothing regarding repetition of vowels or digits.

So students can have ID codes such as AA111 or EE555 or II999 and still their ID codes will not be same.

So, in disguise, the ques is asking - how many such unique ID codes (which doesn't mean unique letters) are possible in total?

which is 5 * 5 * 10 * 10 * 10 = 25000
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Given: ID codes issued by XYZ high school have 5 characters. First 2 characters of the ID codes are vowels and next 3 characters are digits.

Asked: If ID code of no two students is same, then what is the maximum number of students in XYZ high school?

Total number of ID cards = Maximum number of students = 5*5*10*10*10 = 25000

IMO C
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