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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 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
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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?). –  André Nicolas Nov 11 '11 at 7:55
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2 Answers 2

up vote 1 down vote accepted

Consider cases

$x57$ where $x$ is from 1-9 but not $5$ ($9-1=8$ possiblities)

$57x$ where $x$ is from 0-9 but not $5$ ($10-1=9 $possiblities)

$xyz$ where $x$ is from 1-9 but not not 5 and $y,z$ are 0-9 but not 5 ($8\times 9\times 9=648$ possiblities)

So there are $665$ such numbers.

EDIT

So I just read that you require only even numbers.

You have...

$57x$ , here x can only be $0,2,4,6,8$ which is just $5$ possibilites.

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.

(Or alternatively, To correct my previous solution, count out the odd numbers from each case, so you have $665-8-4-8*9*4 =365$)

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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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