2
$\begingroup$

So I am not entirely sure I have approach the question correctly and at first I thought it to be pretty straight forward but there is something in the wording of the question that make me think there is more to it.

A safety lock on a brief case has a keypad. It is unlocked by pressing four digits each in the range $0$ to $9$ inclusive in succession. Mechanical restrictions imply that no single digit features more than once in the code and the order of pressing of the four digits is irrelevant.

How many distinct combinations can there be?

A thief has stolen the brief case and is attempting to open it by trying different combinations. It takes $5$ seconds for him to key in one combination. If he does this systematically using a prepared list of numbers,

  1. What is the maximum time elapsed before he must succeed?
  2. What is the time elapsed by which he has a $50\%$ chance of success?

So my solutions are as follows:

For the number of combinations we are essentially looking at $10\times 9 \times 8=720$ possible combinations.

For 1. I believe it would take him an hour to crack open the case as, the last set of numbers of the prepared could be the combination he looking for. So my calculation $720\times5=3600~\text{s}=1~\text{hr}$

Question 2 is the part that tripping me up as, he has $50\%$ chance of success, so originally I thought well this is half the number of combinations which mean that he would then only take $30$ min to crack the case.

But thinking a little more I can't help but think this is wrong, but I don't really have a solid justification way it more a intuition telling me something not quite right, if I am incorrect could some maybe give me a nudge in the right direction.

$\endgroup$
5
  • $\begingroup$ Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. $\endgroup$
    – Shaun
    Sep 27, 2018 at 11:09
  • 1
    $\begingroup$ You are choosing 4 digits, but multiplying 3 numbers to find the number of combinations. The problem says that order is irrelevant, but you are not dividing by the number of ways to order the chosen digits. For part 2, your intuition was correct. Trying 50% of the combinations would give him a 50% chance of opening the lock. $\endgroup$ Sep 27, 2018 at 11:12
  • $\begingroup$ Yes I miss understand the question but now, have seen where I have been going wrong, as mentioned in previous comment I always get the word order dos'nt matter the wrong way round when looking at permutation and combinations. $\endgroup$
    – james2018
    Sep 27, 2018 at 11:31
  • $\begingroup$ I was thinking if I were using a combination lock then the order itself would matter as in on a combination lock 10-17-23 would be accept the same as 23-17-10. I miss read order dose not matter. $\endgroup$
    – james2018
    Sep 27, 2018 at 11:34
  • $\begingroup$ Now if the problem for part 2 said he was trying random combinations each time, he would need 146 attempts before having at least a 50% chance of guessing correctly. $\endgroup$ Sep 27, 2018 at 11:37

1 Answer 1

4
$\begingroup$

First, note that the combination is four digits, not three. The number of different permutations of four digits (where the order of the digits matters) is

$10 \times 9 \times 8 \times 7 = 5040$

But we know the order in which the digits are entered does not matter. So the combination $1234$ can be entered as $1-3-4-2$ or as $4-2-1-3$ or as $3-2-4-1$ etc. etc. In fact there are $4 \times 3 \times 2 \times 1=24$ permutations (different orderings) for each combination. So the number of different combinations is

$\frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = \frac{5040}{24}=210$

Multiply this by $5$ seconds to find the maximum time that the thief will take to find the combination. If each combination is equally likely then he has a $50\%$ chance of success after half of this maximum time.

$\endgroup$
1
  • $\begingroup$ Ah yes I have just realised my self, what I have done. And I too now have the same ans, many thanks. I always get the order does not matter thing the wrong way round, once again many thanks. $\endgroup$
    – james2018
    Sep 27, 2018 at 11:28

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .