1. ## Lock Combos

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?

2. Originally Posted by magentarita
A combination lock uses three digits from 0-9. If the product of the digits are even, how many lock combos are possible?
If the product is even, then at least one of the digits must be even.
There are $\displaystyle 10^3$ possible combinations. WHY?
There are $\displaystyle 5^3$ possible combinations having only odd digits. WHY?

3. ## sorry...

Originally Posted by Plato
If the product is even, then at least one of the digits must be even.
There are $\displaystyle 10^3$ possible combinations. WHY?
There are $\displaystyle 5^3$ possible combinations having only odd digits. WHY?
I don't get it.

4. Originally Posted by magentarita
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?
$\displaystyle \text{\# of combinations where order matters: }^{\text{10}} \text{P}_\text{3} = 720$

$\displaystyle \text{We have 5 odd numbers i}\text{.e}\text{. 1,3,5,7,9}$

$\displaystyle \text{\# of combos of only odd numbers }^\text{5} \text{P}_\text{3} = 60$

$\displaystyle \text{Hence \# of even combos: }^{\text{10}} \text{P}_\text{3} \text{ - }^\text{5} \text{P}_\text{3} = 720 - 60 = 660.$

5. Originally Posted by Math's-only-a-game
$\displaystyle \text{\# of combinations where order matters: }^{\text{10}} \text{P}_\text{3} = 720$

$\displaystyle \text{We have 5 odd numbers i}\text{.e}\text{. 1,3,5,7,9}$

$\displaystyle \text{\# of combos of only odd numbers }^\text{5} \text{P}_\text{3} = 60$

$\displaystyle \text{Hence \# of even combos: }^{\text{10}} \text{P}_\text{3} \text{ - }^\text{5} \text{P}_\text{3} = 720 - 60 = 660.$
Math's-only-a-game, Why do you assume that the three numbers are all different?
Surely $\displaystyle 525$ could be the code to a combination lock.

6. Originally Posted by Plato
Math's-only-a-game, Why do you assume that the three numbers are all different?
Surely $\displaystyle 525$ could be the code to a combination lock.
Yep, you are right. I'll try again. Thanks for your correction.

7. ## ok

I want to thank all who took time to reply, especially Plato.

8. Originally Posted by magentarita
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?
$\displaystyle \text{Basics:}$

$\displaystyle \text{If the order doesn't matter, it is a Combination}\text{.}$

$\displaystyle \text{If the order does matter it is a Permutation}\text{.}$

$\displaystyle \text{In this case it is a permutation where repetitions are allowed}\text{.}$

$\displaystyle \text{Permutations with Repetition:}$

$\displaystyle \text{If you have n things to choose from, and you choose r of them, }$
$\displaystyle \text{then the permutations are:}$

$\displaystyle \text{n x n x }...\text{ (r times) = nr}$

$\displaystyle \text{(Because there are n possibilities for the first choice, }$

$\displaystyle \text{THEN there are n possibilites for the second choice, and so on}\text{.)}$

$\displaystyle \text{Hence as plato said there are }10^3 \text{lock permutations and as there are 5 odds digits }$
$\displaystyle \text{we have }5^3 \text{odd digit permutations}\text{.}$

$\displaystyle \text{Answer: }10^3 - 5^3 = 1000 - 125 = \text{875 permutations having even digits }$
$\displaystyle \text{and hence their product is even}\text{.}$