# combination lock (question)

• May 16th 2009, 12:17 PM
brentwoodbc
combination lock (question)
The dail on a 3 number combination lock contains markings to represent the numbers 0-59. How many combination's are possible in each case?

a. the first and second numbers must be different, and the second and third numbers must be different.

b. the first and second numbers must differ by atleast 3.

I dont know how to do these.

is it.... $\displaystyle 3^{60}=4.24*10^{28}$ divided by something.........????
• May 16th 2009, 01:54 PM
aidan
Quote:

Originally Posted by brentwoodbc
The dail on a 3 number combination lock contains markings to represent the numbers 0-59. How many combination's are possible in each case?

a. the first and second numbers must be different, and the second and third numbers must be different.

b. the first and second numbers must differ by atleast 3.

I dont know how to do these.

is it.... $\displaystyle 3^{60}=4.24*10^{28}$ divided by something.........????

If there were no restrictions on the 2nd and 3rd number then you could have 60 choices for the 1st, 2nd and 3rd.

$\displaystyle 60 \times 60 \times 60 = 216000$ which is $\displaystyle 60^3$

or restated

$\displaystyle 3^{11} < 60^3 < 3^{12}$

That is no where in the range of $\displaystyle 3^{60}$

If the first number were 3, then the 2nd number could not be 1,2,3,4, or 5: you have 55 choices for the 2nd number on your lock.
• May 16th 2009, 02:43 PM
brentwoodbc
thank you so
a= 60*59*59=208860(Evilgrin)
and
b=60*55*60=198000? book says 198360? I get it though thanks.
its a little tricky with the differ by three thing.
• May 17th 2009, 06:07 AM
Plato
Quote:

Originally Posted by brentwoodbc
b=60*55*60=198000? book says 198360? I get it though thanks.

If the first setting is 0 then second cannot be 0,1,2.
Therefore, there are 57 choices for the second setting and 60 for the third.
The same can be said if the first is 59.

If the first setting is 1 then second cannot be 0,1,2,3.
Therefore, there are 56 choices for the second setting and 60 for the third.
The same can be said if the first is 58.

For any number other than 0,1,58,59 all 56 of them, we exclude 5 possiblies of the second setting.
That means we can choose 55 values for the second and 60 for the third.

$\displaystyle (2)(57+56)(60)+(55)(56)(60)=198360$. The textbook is correct.
• May 18th 2009, 03:48 PM
brentwoodbc
Quote:

Originally Posted by Plato
If the first setting is 0 then second cannot be 0,1,2.
Therefore, there are 57 choices for the second setting and 60 for the third.
The same can be said if the first is 59.

If the first setting is 1 then second cannot be 0,1,2,3.
Therefore, there are 56 choices for the second setting and 60 for the third.
The same can be said if the first is 58.

For any number other than 0,1,58,59 all 56 of them, we exclude 5 possiblies of the second setting.
That means we can choose 55 values for the second and 60 for the third.

$\displaystyle (2)(57+56)(60)+(55)(56)(60)=198360$. The textbook is correct.

how is it 2*(57+56)*60 I dot understand that part totally.
• May 18th 2009, 03:56 PM
brentwoodbc
you also said if its 1 it cant be 0,1,2,3 , but what about if its 0?

wouldnt it be 2*(57+57)*60 ? I dont get this.
• May 18th 2009, 04:01 PM
Plato
Quote:

Originally Posted by Plato
If the first setting is 0 then second cannot be 0,1,2.
Therefore, there are 57 choices for the second setting and 60 for the third.
The same can be said if the first is 59.

If the first setting is 1 then second cannot be 0,1,2,3.
Therefore, there are 56 choices for the second setting and 60 for the third.
The same can be said if the first is 58.

For any number other than 0,1,58,59 all 56 of them, we exclude 5 possiblies of the second setting.
That means we can choose 55 values for the second and 60 for the third.

$\displaystyle (2)(57+56)(60)+(55)(56)(60)=198360$. The textbook is correct.

Quote:

Originally Posted by brentwoodbc
you also said if its 1 it cant be 0,1,2,3 , but what about if its 0?
wouldnt it be 2*(57+57)*60 ? I dont get this.

Do you actually read the replies?
• May 18th 2009, 04:08 PM
brentwoodbc
Quote:

Originally Posted by Plato
If the first setting is 0 then second cannot be 0,1,2.
Therefore, there are 57 choices for the second setting and 60 for the third.
The same can be said if the first is 59.

If the first setting is 1 then second cannot be 0,1,2,3.
Therefore, there are 56 choices for the second setting and 60 for the third.
The same can be said if the first is 58.

For any number other than 0,1,58,59 all 56 of them, we exclude 5 possiblies of the second setting.
That means we can choose 55 values for the second and 60 for the third.

$\displaystyle (2)(57+56)(60)+(55)(56)(60)=198360$. The textbook is correct.

(2)(57+56)(60)

I just need clarification on this part
2= number of digits where you have 57/56 choices?
(57+56) = why do you add?

is it 2 choices (times) where you have 57 and 56 choices multiplied by 60?