
A letter lock consists of $4$ rings each marked with $10$ different letters. Then, the number of ways of making unsuccessful attempts to open the lock is equal to $k$, then the value of $\dfrac{k}{9999}$ is?
Answer
458.1k+ views
Hint: In this problem we need to calculate the value of $\dfrac{k}{9999}$, where $k$ is the number of unsuccessful attempts to open a lock which consists of $4$ rings each marked with $10$ different letters. So here we will first calculate the number of possible $4$ lettered words that can be formed by using the $10$ different letters. These are the all-possible attempts we can do to open the lock. Out of them we must have a one $4$ lettered word with which we can open the lock. So, the number of unsuccessful attempts will be one less than the total number of possible attempts. From this we can calculate the total number of unsuccessful attempts which is our $k$ value. From this we can easily calculate the required value.
Complete step by step answer:
Given that, A letter lock consists of $4$ rings each marked with $10$ different letters.
The total number of possible attempts we can do to open the lock by using $10$ different letters with $4$ rings is given by
$T={{10}^{4}}$
Applying the exponential formula ${{a}^{n}}=a\times a\times a\times .....\text{ n times}$ and simplifying the value, then we will get
$\begin{align}
& T=10\times 10\times 10\times 10 \\
& \Rightarrow T=10000 \\
\end{align}$
In the above $10000$ attempts we must have a one sign successful attempt, then the total number of unsuccessful attempts is given by
$\begin{align}
& U=10000-1 \\
& \Rightarrow U=9999 \\
\end{align}$
Hence the total number of unsuccessful attempts is equal to $9999$.
But in the problem, they have mentioned that the total number of unsuccessful attempts is equal to $k$, so we can write
$k=9999$
Dividing the above equation with $9999$ on both sides, then we will get
$\begin{align}
& \dfrac{k}{9999}=\dfrac{9999}{9999} \\
& \Rightarrow \dfrac{k}{9999}=1 \\
\end{align}$
Hence the value of $\dfrac{k}{9999}$ is $1$.
Note: In this problem they have not mentioned about the repetition of letters while attempting to open the lock. So, we have considered the repetition is allowed. If they have mentioned that the repetition is not allowed then the value of the total number of attempts to open the lock will be changed and they by our answer will also change.
Complete step by step answer:
Given that, A letter lock consists of $4$ rings each marked with $10$ different letters.
The total number of possible attempts we can do to open the lock by using $10$ different letters with $4$ rings is given by
$T={{10}^{4}}$
Applying the exponential formula ${{a}^{n}}=a\times a\times a\times .....\text{ n times}$ and simplifying the value, then we will get
$\begin{align}
& T=10\times 10\times 10\times 10 \\
& \Rightarrow T=10000 \\
\end{align}$
In the above $10000$ attempts we must have a one sign successful attempt, then the total number of unsuccessful attempts is given by
$\begin{align}
& U=10000-1 \\
& \Rightarrow U=9999 \\
\end{align}$
Hence the total number of unsuccessful attempts is equal to $9999$.
But in the problem, they have mentioned that the total number of unsuccessful attempts is equal to $k$, so we can write
$k=9999$
Dividing the above equation with $9999$ on both sides, then we will get
$\begin{align}
& \dfrac{k}{9999}=\dfrac{9999}{9999} \\
& \Rightarrow \dfrac{k}{9999}=1 \\
\end{align}$
Hence the value of $\dfrac{k}{9999}$ is $1$.
Note: In this problem they have not mentioned about the repetition of letters while attempting to open the lock. So, we have considered the repetition is allowed. If they have mentioned that the repetition is not allowed then the value of the total number of attempts to open the lock will be changed and they by our answer will also change.
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Trending doubts
1 ton equals to A 100 kg B 1000 kg C 10 kg D 10000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

Explain zero factorial class 11 maths CBSE
