Question

The number of all possible three digit even numbers which can be formed with the condition that if $5$ is one of the digits, then $7$ is the next digit, is equal to

• A. $5$
• B. $325$
• C. $345$
• D. $365$

Answer 1

Answer: (D)

If $5$ is at hundreds place, $7$ will be at tens place and ones place will have $0,2,4,6$ or $8$ . So, $5$ numbers are possible.
If $5$ is at tens place then, $7$ will be at ones place but it won’t be an even number.
If $5$ is at ones place, then it won’t be an even number. So, $5$ can only be at hundreds place
$\underline{5} \underline{7} \underline{x} \rightarrow x=0,2,4,6,8 \rightarrow 5$ numbers
$\underline{8} \times \underline{9} \times \underline{5}=360$ numbers
(any number except $0,5$ at hundreds place, any number except $5$ at tens place and $0,2,4,6,8$ at ones place)
Total numbers possible $=5+360=365$