### Character code puzzle

There is a 4-character code, with 2 of them being letters and the other 2 being numbers.

How many maximum attempts would be necessary to find the correct code? Note that the code is case-sensitive.

The maximum number of attempts required are 16,22,400

There are 52 possible letters - a to z and A to Z, and 10 possible numbers - 0 to 9. Now, 4 characters - 2 letters and 2 numbers, can be selected in 52*52*10*10 ways. These 4 characters can be arranged in 4C2 i.e. 6 different ways - the number of unique patterns that can be formed by lining up 4 objects of which 2 are distinguished one way (i.e. they must be letters) and the other 2 are distinguished another way (i.e. they must be numbers).

Consider an example : Let's assume that @ represents letter and # represents number. the 6 possible ways of arranging them are : @@##, @#@#, @##@, #@@#, #@#@, ##@@

= 52*52*10*10*6
= 16,22,400 attempts
= 1.6 million approx.

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