no clue: max time to find locker combination

[tsh44]

Ok the locks on gym lockers have dials with numbers 0-39. Each locker combo consists of 3 numbers. A girl can't remember any of her combo numbers but remembers that 2 of them are exactly the same. If she can try one possibility every ten seocnds what is the maximum amount of time that it would take her to find the right combo?

I tried 40x 39 x 39 but it didn't work out. I would appreciate your help.

 

[tkhunny]

What didn't work out?

Order matters for a combination.

 

[tsh44]

my teacher said order matters for a permutation

the answer comes out to be 13 hrs accoridng to the book but I did not get that when I did 40 x 39 x 39.

 

[stapel]

my teacher said order matters for a permutation

This is correct.

the answer comes out to be 13 hrs accoridng to the book

This also is correct.

but I did not get that when I did 40 x 39 x 39.

What is the reasoning behind "40 × 39 × 39"? Where are you accounting for the time for each attempt? Where are you converting from "total seconds" to "hours"?

Please reply with all of your work and reasoning. Thank you.

Eliz.

 

[pka]

You textbook is correct. Study the following.
\(\displaystyle \L
(3 \cdot 40 \cdot 39) = 4680\quad \& \quad \frac{{13 \cdot 60^2 }}{{10}} = 4680\).

How many seconds in 13 hours?
How many ways can we arrange the three numbers, 8-13-8?

 

[tsh44]

pka, i don't understand the logic in 3x40x39. There are 46800 seconds in 13 hours.

 

[pka]

Lets say that the lock’s combination really consists two 8’s and one 13.
Now then the actual combination could be one of: 8-8-13, 8-13-8, or 13-8-8.
That is three ways to order <8,8,13>.

Now how many multisets <a,a,b> are there?
There are 40 ways to select the a and 39 ways to select the b.
There are 3 ways the arrange that multiset into a lock combination.

Therefore we get 3*40*39 possible lock combinations.

 

[soroban]

Hello, tsh44!

The locks on gym lockers have dials with numbers 0 - 39.
Each locker combo consists of 3 numbers. A girl can't remember any of her combo numbers
but remembers that 2 of them are exactly the same.
If she can try one possibility every ten seocnds,
what is the maximum amount of time that it would take her to find the right combo?

I tried 40 x 39 x 39 but it didn't work out. . . . It wouldn't.

If the numbers are $\displaystyle ABB$ (the second and third numbers are the same),
$\displaystyle \;\;$there are $\displaystyle 40$ choices for the $\displaystyle A$
$\displaystyle \;\;$there are $\displaystyle 39$ choices for the $\displaystyle B$
$\displaystyle \;\;$and one choice for the second $\displaystyle B.$
Hence, there are $\displaystyle 40\,\times\,39\,\times\,1$ choices for $\displaystyle ABB.$

But the combination could be $\displaystyle \,ABB,\;BAB,\,$ or $\displaystyle BBA$.
So there are: $\displaystyle \,3\,\times\,40\,\times\,39\:=\:4680$ possible combinations.

At 10 seconds per combination, it will take:
$\displaystyle \;\;10\,\times\,4680\:=\: 46,800$ seconds $\displaystyle \:=\:780\,$ minutes $\displaystyle \:=\:13$ hours.

Edit: Too fast for me, pka!