
A number lock has 4 dials, each dial has the digits 0, 1, 2, …….., 9. What are the maximum unsuccessful attempts to open the lock?
Answer
606.3k+ views
Hint:Find all the patterns of numbers that can be formed by using the digits 0, 1, 2, …….., 9 in four blank places. Each digit can be repeated at the four places, so, apply the formula: ${{n}^{4}}$ to determine the total attempts. Here, ‘n’ is the number of digits available to form a lock pattern and index ‘4’ is the number of places in which these digits can be filled. Subtract 1 from the total attempts obtained, to get the total number of unsuccessful attempts because there is only one successful attempt.
Complete step-by-step answer:
We have been provided a number lock which has 4 dials, that means the lock is formed by using any four digits among 0, 1, 2, …….., 9. We know that, in a lock pattern numbers can be repeated.
Therefore, we can fill all the dials with any of the ten digits 0, 1, 2, …….., 9.
Number of ways to fill 1st dial = 10
Number of ways to fill 2nd dial = 10
Number of ways to fill 3rd dial = 10
Number of ways to fill 4th dial = 10
Since, we have to fill all the four dials one by one, therefore, the total number of patterns that can be formed $=10\times 10\times 10\times 10={{10}^{4}}$.
Now, we know that among ${{10}^{4}}$ ways only one pattern will be correct. Therefore, the remaining patterns are incorrect. We have to find the maximum number of unsuccessful attempts, so have to assume that the correct pattern occurs at last.
Hence, maximum number of unsuccessful
$\begin{align}
& ={{10}^{4}}-1 \\
& =10000-1 \\
& =9999 \\
\end{align}$
Note: One may note that, it may be possible, we can get the correct pattern in the first attempt only. So the number of unsuccessful attempts will be 0. In any of the ${{10}^{4}}$ attempts we can get the correct pattern. If this is the situation, then we would not be able to solve the problem. That is why it is said in the question that we have to find the maximum number of unsuccessful attempts. That means the only successful attempt should be considered at last. So, do not get confused in this situation. First find the total attempts and then subtract 1 from it to get the maximum number of unsuccessful attempts.
Complete step-by-step answer:
We have been provided a number lock which has 4 dials, that means the lock is formed by using any four digits among 0, 1, 2, …….., 9. We know that, in a lock pattern numbers can be repeated.
Therefore, we can fill all the dials with any of the ten digits 0, 1, 2, …….., 9.
Number of ways to fill 1st dial = 10
Number of ways to fill 2nd dial = 10
Number of ways to fill 3rd dial = 10
Number of ways to fill 4th dial = 10
Since, we have to fill all the four dials one by one, therefore, the total number of patterns that can be formed $=10\times 10\times 10\times 10={{10}^{4}}$.
Now, we know that among ${{10}^{4}}$ ways only one pattern will be correct. Therefore, the remaining patterns are incorrect. We have to find the maximum number of unsuccessful attempts, so have to assume that the correct pattern occurs at last.
Hence, maximum number of unsuccessful
$\begin{align}
& ={{10}^{4}}-1 \\
& =10000-1 \\
& =9999 \\
\end{align}$
Note: One may note that, it may be possible, we can get the correct pattern in the first attempt only. So the number of unsuccessful attempts will be 0. In any of the ${{10}^{4}}$ attempts we can get the correct pattern. If this is the situation, then we would not be able to solve the problem. That is why it is said in the question that we have to find the maximum number of unsuccessful attempts. That means the only successful attempt should be considered at last. So, do not get confused in this situation. First find the total attempts and then subtract 1 from it to get the maximum number of unsuccessful attempts.
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