1. ## How many combinations..?

Hi!

i am having this difficulty at the moment. My Dad has forgotten the combination to his bicycle lock... he knows (believes) it is a combination of 44670. We have tried the more obvious ones such as 07644, and moving the 0 around. I am wondering how many combinations of 44760 are there?

if i've got nothing to do i i'll try them and see. What has stopped me from working it out is the 4 appearing twice, i don't know how that affects these kind of things.

also, just as kind of "i want to know just so i don't try it" kind of thing, if the combination lock has 5 rows of ten numbers, how many combinations of those numbers can there be? we thought it was 10x10x10x10x10 = 100000

Anyway, thanks for any help!

James

2. Originally Posted by jiminwatford
Hi!

i am having this difficulty at the moment. My Dad has forgotten the combination to his bicycle lock... he knows (believes) it is a combination of 44670. We have tried the more obvious ones such as 07644, and moving the 0 around. I am wondering how many combinations of 44760 are there?
${5 \choose 2} = 10$ (Assuming the there is one digit doubled, in this case the 4. Note the 4's need not be next to each other for this number of arrangements)

(I suggest writing all the combinations down first on a piece of paper then trying them physically.)

Originally Posted by jiminwatford
if i've got nothing to do i i'll try them and see. What has stopped me from working it out is the 4 appearing twice, i don't know how that affects these kind of things.
Are you quite certain the 4 appears twice? Maybe you should try doubling the 6 or the 7, etc in the combination instead of the 4.

Originally Posted by jiminwatford
also, just as kind of "i want to know just so i don't try it" kind of thing, if the combination lock has 5 rows of ten numbers, how many combinations of those numbers can there be? we thought it was 10x10x10x10x10 = 100000
$10^5 = 100 000$ combinations.

Originally Posted by jiminwatford
Anyway, thanks for any help!

James
Hope I could help

3. hi, the four would be the only one doubled as he used the numbers from his birthday and had to use a 0 for the extra line.

i'm not sure i follow. are you saying there is only 10 combinations those numbers can be arranged in? I must have misunderstood, as when i keep the 4s together there is 24 combinations

thanks
James

4. Originally Posted by jiminwatford
hi, the four would be the only one doubled as he used the numbers from his birthday and had to use a 0 for the extra line.

i'm not sure i follow. are you saying there is only 10 combinations those numbers can be arranged in? I must have misunderstood, as when i keep the 4s together there is 24 combinations

thanks
James
I may have made a mistake in my logic somewhere...

Lets rather write it out...

44670
44760
44607
44706
44076
44067

04467
04476
64470
64407
74460
74406

06447
60447
07446
70446
67440
76440

06744
07644
60744
67044
70644
76044

Which is 24 as you calculated. I'll rethink my calculations.

========

What I would try:

44076
44706
44607
44067

5. Oh i get it.

$4!$ gives us the answer because the fourth digit is the same as the fifth, giving us four distinct digits and not 5.

6. Hello, James!

My Dad has forgotten the combination to his bicycle lock.
He believes it is a combination of 44670.
I am wondering how many combinations of 44760 there are.

Although it is called a "combination" lock, the order of the digits is significant.

If those were five different digits, there would be $5! = 120$ possible sequences.

In each of the 120 sequences, the two 4's can be switched
. . without creating a new sequence.

Hence, our answer is twice as large as it should be.

Therefore, the number of possible "combinations" is: . $\frac{120}{2} \:=\:60$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Here's another approach . . .

$\text{We have five spaces to fill: }\;\_\;\;\_\;\;\_\;\;\_\;\;\_$

Pick two spaces for the two 4's . . . There are: . ${5\choose2} \:=\:10$ choices.

The other three digits {0,6,7} can be placed in: . $3! \:=\:6$ ways.

Therefore, there are: . $10 \times 6 \:=\:60$ possible "combinations".

7. Originally Posted by Soroban
Hello, James!

Although it is called a "combination" lock, the order of the digits is significant.

If those were five different digits, there would be $5! = 120$ possible sequences.

In each of the 120 sequences, the two 4's can be switched
. . without creating a new sequence.

Hence, our answer is twice as large as it should be.

Therefore, the number of possible sequences is: . $\frac{120}{2} \:=\:60$

Soroban, if he says he knows for sure the two fours should be next to each other, wouldn't that change it? (Your solution assumes the fours need not be next to each other, am I right?)

8. Hello, janvdl!

In that case, we have four "numbers" to arrange: . $\{0,\;\boxed{44},\:6,\:7\}$

. . Therefore, there are: . $4! = 24$ "combinations".

9. Hi, thanks!

you are right, the 4s don't need to be together. i had worked out the 24 possibilities when keeping the 4s together and none of them worked, so i have to try them seperated. It is unlikely they are actually seperated as my Dad's birthday is in 19'44' so would probably have not seperated but... we need to try all things

i just wish i hade the patience to try all 100,000 options!

thanks

10. Originally Posted by jiminwatford
Hi, thanks!

you are right, the 4s don't need to be together. i had worked out the 24 possibilities when keeping the 4s together and none of them worked, so i have to try them seperated. It is unlikely they are actually seperated as my Dad's birthday is in 19'44' so would probably have not seperated but... we need to try all things

i just wish i hade the patience to try all 100,000 options!

thanks
Erm... Why not just grind the thing off? its much much easier, unless that lock is worth \$1000