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

Hint: 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.