A combination lock uses three digits from 0-9. If the product of the digits are even, how many lock combos are possible?
Do I use 3C9?
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\text{\# of combinations where order matters: }^{\text{10}} \text{P}_\text{3} = 720
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\text{We have 5 odd numbers i}\text{.e}\text{. 1,3,5,7,9}
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\text{\# of combos of only odd numbers }^\text{5} \text{P}_\text{3} = 60
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\text{Hence \# of even combos: }^{\text{10}} \text{P}_\text{3} \text{ - }^\text{5} \text{P}_\text{3} = 720 - 60 = 660.
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\text{Basics:}
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\text{If the order doesn't matter, it is a Combination}\text{.}
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\text{If the order does matter it is a Permutation}\text{.}
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\text{In this case it is a permutation where repetitions are allowed}\text{.}
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\text{Permutations with Repetition:}
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\text{If you have n things to choose from, and you choose r of them, }
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\text{then the permutations are:}
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\text{n x n x }...\text{ (r times) = nr}
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\text{(Because there are n possibilities for the first choice, }
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\text{THEN there are n possibilites for the second choice, and so on}\text{.)}
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\text{Hence as plato said there are }10^3 \text{lock permutations and as there are 5 odds digits }
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\text{we have }5^3 \text{odd digit permutations}\text{.}
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\text{Answer: }10^3 - 5^3 = 1000 - 125 = \text{875 permutations having even digits }
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\text{and hence their product is even}\text{.}
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