chetan2u wrote:

How many 3-digit even numbers are possible such that if one of the digits is 5, the next/succeeding digit to it should be 7?

A) 5

B) 305

C) 365

D) 405

E) 495

Since the number should be even, the last place can be taken by 0,2,4,6 or 8.

Since 5 has to be followed by 7, 5 can't be the tens digit.

Knowing this information we can solve this problem:

Case 1: When 5 is present at the hundreds place:

The number will be 57_. Now _ can be filled in 5 ways - 0,2,4,6 or 8.

Thus total ways=5Case 2: When 5 is not present at the hundreds place:

The number will have 8 digits at the hundreds place (except 0 & 5).

The number will have 9 digits at the tens place (except 5).

The number will have 5 digits at the units place.

So total ways=8*9*5=360Hence, total ways=360+5=365
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