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

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 your three digit number is $abc$ with $c$ even, you either have $57c$ or $abc$ with $a \neq 5 \neq b$, am I right? Or did I misunderstand your question? – Patrick Da Silva Nov 11 '11 at 7:50
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?). – André Nicolas Nov 11 '11 at 7:55

2 Answers

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? – Arthur Fischer Jul 28 at 10:27