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How many three digits even numbers can we form such that if one of digit is $5$ the following digit must be $ 7$?

I need some ideas on how to proceed on this problem.

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    If the beginning is $5$, then next is $7$, now how many ways for last? If the beginning is not $5$ ($8$ ways), the next is anything but $5$ (how many ways?) and the last is even (how many ways?).2011-11-11

2 Answers 2

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You have two different kinds of such three-digit even numbers.

  • $57x$, where $x$ can only be $0,2,4,6,8$ which is just $5$ possibilities.

  • For the remaining, you count all even three-digit numbers with no $5$ in them. This will be $8\times 9\times 5 = 360$

So you have $365$ possibilities.

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required number of ways = number of ways is number of ways of forming a 3 digit even number(10.10.5)- number of even numbers in which 5 is the first digit and 7 is not the second digit(1.9.5) = 455

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    10 choices for the first digit? Really?2015-07-28