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Given a $3\times4$ matrix keypad, each key encoded onto a unique index on a 12 bit string (0000-0000-0000), the maximum combinations are $2^{12}=4096$.
However, $12$ available keys have a maximum possible combination of $12!$.
Obviously, it's pretty hard to map $12!$ values into a space of $4096$ possible combinations.
What am I missing here? I've been trying to figure it out for an hour.
1 Answer
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If you had 12 different symbols to put in 12 slots, you would have $12!$ different ways to do this (assuming each symbol is used once). However you are putting $0$s and $1$s into the slots--two choices per slot for a total of $2^{12}$ ways to do it.
