Question

How many even numbers greater than $300$ can be formed with the digits $1$, $2$, $3$, $4$, $5$ if repetition of digits in a numbers is not allowed.

Answer 1

First find the $4$digit even number possibilities,$3$ digits even number possibilities and 5 digits even number possibilities and find last digit possibility in both the cases. Now, take the sum of all the possibilities that will be the even number greater than $300$.

Complete Step-by-step Solution
Number can be $3$, $4$ or $5$ digit.
$5$digit even no’s:
Last digit must be either $2$ or $4$, so $2$ choices.
First $4$digit even no: possible = $4! \times 2 = 48$
$4$digit even no:
Last digit again must be either $2$ or $4$. So $2$ choice i.e., ${}^4{P_3} \times 2 = 48$.
$3$ digits even no’s
Last digit again must be either $2$ or $4$. So $2$ choice i.e.,${}^4{P_2} = 24.$
But out of these $24$,$3 -$digit no: some will be less than $300$, the ones starting with $1$ or $2$.
These are $9$such no’s ($1 \times 2$ and $3$ options for the ${2^{nd}}$digit $x$, so 3$$$$ + \left( {1 \times 4} \right) and $3$options for $x$,
So $\left( {3 + 2 \times 4} \right)$ and $3$ options for $x.$
\therefore 24 - 9$$$$ = 15
\therefore $$$$48 + 48 + 15 = 111.

$\therefore$ Option (B) is the correct option.

Note:
In these kinds of problems, one needs to focus on the probability take the sum of all the possibilities that will be the even or odd number greater than$n$. $n$ can be any given number. We need to be careful at calculation steps and while writing factorials.