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Question also mentions "alphabetical order" . 5c2 /5c3 or 6c2/6c3 will give all possible combination for ex -: 'CBA' . But thats not a valid code . I calculated it manually till 4 and guessed 6 to be the answer as 4 gave only 10 codes .

How to tackle this in exam ?
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Assuming n be the number of alphabets we have 2(n-1)+2(n-2) code depending upon they being in alphabetical order.
Now n=8.5 with this. so we need 9 letters. Hence answer is E.
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Question also mentions "alphabetical order" . 5c2 /5c3 or 6c2/6c3 will give all possible combination for ex -: 'CBA' . But thats not a valid code . I calculated it manually till 4 and guessed 6 to be the answer as 4 gave only 10 codes .

How to tackle this in exam ?

Combination is the most efficient way to solve this. 'CBA' will be possible case if we use permutation instead of combination; combination has less number of total cases than permutation and we require less number of cases in this problem in order to rule out 'CBA'.
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