Combinatorics problem

Question

Can You please help with this problem:

A gate code requires a four-digit code for entry. Still, for security reasons, it always randomly assigns the digits to key positions (10 keys) so that all possible assignments have the same probability. If our code consists of four different numbers (0-9), what is the probability that we leave fingerprints in the same positions during two logins?

I appreciate any help you can provide.

Answer 1

[Tutor's NOTE: Read the problem carefully and get to the Math core on the first reading. It's especially important on a timed test, like the Digital SAT]

Observation 1: There is "four-digit code" of independent digit positions, with all the codes being different security values. Accordingly, we are talking Permutations (not Combinations).

Observation 2: There is "in the same positions" (not, in a given position), which is not given/defined. Accordingly, the first login defines the favorable outcome (therefore, not a try). Then, the second login's fingerprint matching the first fingerprint - is the single try.

Probability = favorable outcomes / all possible outcomes.

Favorable outcomes (single match of a unique code) = 1

All possible outcomes (permutations of 4 independent 10-key positions) = 10⁴

Probability = 1/10⁴

Answer: Probability of leaving fingerprints in the same positions during two logins is 1/10⁴.

Answer 2

It seems more plausible that different digits will be assigned to different keys.

Let's rephrase the problem. There are ten digits and ten keys. At each login, each digit is assigned to a different key. The four digit code corresponds to a subset of size four from the keys, and there are 10 choose 4 such subsets. We leave fingerprints on the four keys in the subset. The order doesn't matter.

The problem is asking for the probability that we leave fingerprints in the same position during two successive logins. That's the same as asking that two randomly selected subsets of size 4 from a set of size 10 match. That's just 1 over 10 choose 4, or 4!*6!/10!

Answer 3

One might think this is a permutation problem, yet that would assume each key entry was different, i.e. no 1221, or 5555. That was not stated. So the number of possible four number pass codes is 10⁴.

The probability of selecting any one of them is 1 / 10⁴,

The probability of selecting any one of them twice in a row is (1 / 10⁴)².

Answer 4

Let's breakdown the poroblem step by step:

1) Number of Possible Assignments:

1. There are 10 key positions, and each position can be assigned any of the 10 digits (0-9).
2. So, the total number of possible assignments is (10^4) (since there are 4 positions).

2) Number of Ways to Choose the Same 4 Positions:

1. For the second login, we want to choose the same 4 positions as the first login.
2. The number of ways to choose 4 positions out of 10 is given by the combination formula:

C ( n , r ) = n ! r ! ( n − r ) ! 

C(10,4)= 10! 4! (10-4)!

= 10! 4! 6!

= 210

3) Probability:

1. The probability of leaving fingerprints in the same positions during two logins is the ratio of the number of favorable outcomes (same 4 positions) to the total number of possible outcomes.

Putting it all together:

1. Probability = Number of favorable outcomes / Total number of possible outcomes

P= (210)/10^4

Therefore, the probability that one leaves fingerprints in the same positions during two logins is approximately 0.021 or 2.1%.

If you have any more questions or need further assistance, feel free to ask!