Find the probability that getting 4 digit number with 1 in the unit place and 2 in the tens place when the numbers 1, 2, 3, 4 and 5 are arranged at random without repeating.

Answer: Option D

The numbers can be written in 5! different ways then n(S) = 5! = 120
Let E be the event that getting 4 digit number with 1 in the unit place and 2 in the tens place when the numbers 1, 2, 3, 4 and 5 are arranged at random.
Since each desired number is ending with 1, there is only 1 way
The Second place(tens) can now be filled with 2
Here also only 1 way to fill
The next place(hundreds) can now be filled by any of the remaining 3 numbers
So, there are 3 ways to filling that place
Then, the first place can now be filled by any of the remaining 2 numbers
So, there are 2 ways to fill
Therefore n(E) = 1 x 1 x 3 x 2 = 6
Now,
$$\begin{array}{rl}& \text{p}\left(\text{E}\right)=\frac{\text{n}\left(\text{E}\right)}{\text{n}\left(\text{S}\right)}\\ & =\frac{6}{120}\\ & =\frac{1}{20}\end{array}$$
Hence the required probability is $$\frac{1}{20}$$