Question

The number of unsuccessful attempts that can be made by a thief to open a number lock having 3 rings in which each rings contains 6 numbers is
A.205
B.200
C.210
D.215

Answer 1

Hint: First, we will find the total number of attempts made for the lock to open by multiplying the three cases. Then to open a number lock there is only one way as every lock has unique code and always open with one number, so the number of successful attempts is 1. So to find the number of unsuccessful attempts a thief can make to open a lock is the difference of total attempts to open the lock and number of successful attempts.
Thus to find the number of unsuccessful attempts a thief can make to open a lock is the difference of total attempts to open the lock and number of successful attempts,

Complete step-by-step answer:
We are given that there are 3 rings and each ring contains 6 numbers.
So, we know that the first ring can be filled in 6 ways as it contains 6 letters, then the second ring will also have 6 numbers so it can be filled in 6 ways and for the third ring will have 6 ways.
Now we will find the total number of attempts made for the lock to open by multiplying the three cases, we get
\[
   \Rightarrow 6 \times 6 \times 6 \\
   \Rightarrow 216 \\
 \]
Since we know that to open a number lock there is only one way as every lock has unique code and always open with one number, so the number of successful attempts is 1.
Thus to find the number of unsuccessful attempts a thief can make to open a lock is the difference of total attempts to open the lock and number of successful attempts, we get
\[
   \Rightarrow 216 - 1 \\
   \Rightarrow 215 \\
 \]
Hence, option D is correct.

Note: In solving this question, students make mistakes by taking a number of attempts are \[6 \times 5 \times 4 = 120\], which is wrong. As each ring has 6 digits on its own so we will select from 6 digits for each ring. The other common mistake is to consider total attempts as the number of unsuccessful attempts, which will lead to a wrong answer. Since we know that there is only one successful attempt to open a lock as every lock has its unique code. As we are given 4 options and none of the options contains the possibility of making a mistake but if they include that will be incorrect.