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.

Solution(By Examveda Team)

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,
$$\eqalign{ & {\text{p}}\left( {\text{E}} \right) = \frac{{{\text{n}}\left( {\text{E}} \right)}}{{{\text{n}}\left( {\text{S}} \right)}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{6}{{120}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{{20}} \cr} $$
Hence the required probability is $$\frac{1}{{20}}$$