permutations: A combination lock has 60 different positions

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xo_caroline_xo

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1. A combination lock has 60 different positions. To open the lock, you move to a certain number in a clockwise direction, then to a number in the counterclockwise direction, and finally to a third number in the clockwise directio. If consecutive numbers in the combination cannot be the same, how many different combinations are there?

For this question, I think the solution is:

60! (60 differnt positions)

= 8.32 X 10^81

I don't know if this is correct.

2) There are 1500 students in Hall C.S.S. Each student requires a lock for a personal locker. The school provides a standard brand of lock for all students. If the locks are to operate the same way as those described in question 1, what is the smallest number of positions that must be in the lock to give each student a unique combination?

I don't know what to do with this question.
 
Re: permutations: A combination lock has 60 different positi

xo_caroline_xo said:
For this question, I think the solution is:

60! (60 differnt positions)
Click to expand...
This might be valid if the combination were required to use all sixty numbers, once each. But is that what the exercise states?

Eliz.
 
It says "consecutive".

60 in the first slot.
59 in the second slot.
59 in the third slot.

60! Suggests 60 values in the combination, not three. That's a pretty serious combination. 60! also suggests no number may be used twice and entirely ignores the word "consecutive".

Read more carefully.
 
tkhunny said:
It says "consecutive".

60 in the first slot.
59 in the second slot.
59 in the third slot.

60! Suggests 60 values in the combination, not three. That's a pretty serious combination. 60! also suggests no number may be used twice and entirely ignores the word "consecutive".

Read more carefully.
Click to expand...

what do i do then?!?!

would i multiply 60*59*58 which equals 205320?!
 
It's 60*59*59, since consecutive numbers cannot be repeated, but the first and third can be.

For the second question, notice that the pattern is n * (n-1) * (n-1). If you need at least 1500 unique combinations, then you need to solve the equation:

n * (n-1) * (n-1) >= 1500

You can do trial-and-error by plugging in small numbers for n and seeing where it becomes larger than 1500, or you can try to solve:

n * (n-1) * (n-1) >= 1500
n * (n-1)^2 >= 1500
= n^3 - 2n^2 + 2n >= 1500
= n^3 - 2n^2 + 2n - 1500 >= 0

I would do trial-and-error :D
 
JohnM said:
It's 60*59*59, since consecutive numbers cannot be repeated, but the first and third can be.

For the second question, notice that the pattern is n * (n-1) * (n-1). If you need at least 1500 unique combinations, then you need to solve the equation:

n * (n-1) * (n-1) >= 1500

You can do trial-and-error by plugging in small numbers for n and seeing where it becomes larger than 1500, or you can try to solve:

n * (n-1) * (n-1) >= 1500
n * (n-1)^2 >= 1500
= n^3 - 2n^2 + 2n >= 1500
= n^3 - 2n^2 + 2n - 1500 >= 0

I would do trial-and-error :D
Click to expand...

wouldnt it like take forever if i did trial and error? :?
 
xo_caroline_xo said:
would i multiply 60*59*58 which equals 205320?!
Click to expand...
You're not paying attention. :cry: The word "consecutive" mean you DO get to reuse numbers, just not right next door. As has been demonstrated, it is 60*59*59. There are no 58s in there. Whatever number you used in the first slot, you get to use again in the third slot.


Pick a number: 40? 40*39*39 = 60840 -- Too big
20? 20*19*19 = 7220 -- Too big
10? 10*9*9 = 810 -- Not big enough
15? 15*14*14 = 2940 -- Too big

There are only a few left to check. It doesn't have to take forever. Just sit down and do it.

Alternatively, you could find the solutions in the Real Numbers, rather than restricting to the Natural Numbers. Out pops n = 12.123, so I would try n = 13 as my favorite candidate. I would also try n = 12, just to prove that it is inadequate.
 
so the first answer is 60*59*59

and the second answer would be;

let x be the number of numbers on the lock

1500=desired number of combinations
1500=(x-1)(x-1)(x)
1500=(x^2-x-x+1)(x)
1500=x^3-2x^2+x
0=x^3-2x^2+x-1500

right?!
 
xo_caroline_xo said:
right?!
Click to expand...
1) Not until you solve it.
2) You need a Whole Number. You cannot make a lock with a fractional number of digits.
 
xo_caroline_xo said:
tkhunny said:
[quote="xo_caroline_xo":2nisyxle]right?!
Click to expand...
1) Not until you solve it.
2) You need a Whole Number. You cannot make a lock with a fractional number of digits.
Click to expand...

can u help me because i dont know what to do next[/quote:2nisyxle]

In the equation:

n * (n-1) * (n-1) >= 1500

start by plugging in an integer for n, and compute the answer. Then keep increasing n until you get an answer that is a little bigger than 1500.

Don't worry - it's not a huge number at all....
 
xo_caroline_xo said:
can u help me because i dont know what to do next
Click to expand...
You're kidding, right? Did you see my post at 2:52? What did I leave out?
 
xo_caroline_xo said:
i dont understand how to do question 2
Click to expand...
Try following the step-by-step instructions, provided earlier:

tkhunny said:
Pick a number: 40? 40*39*39 = 60840 -- Too big
20? 20*19*19 = 7220 -- Too big
10? 10*9*9 = 810 -- Not big enough
15? 15*14*14 = 2940 -- Too big

There are only a few left to check. It doesn't have to take forever. Just sit down and do it.
Click to expand...
So try 11, 12, 13, and 14, until you find the one that works.

Eliz.
 
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