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How many $3$ digit even numbers can be formed by using digits $1,2,3,4,5,6,7$, if no digits are repeated?

ATTEMPT

There are three places to be filled in _ _ _ I wrote it like this

_ _ $2$

_ _ $4$

_ _ $6$

Now each of the two blanks can be filled in $P(6,2)$ ways. So adding results of three cases i have $3$. $P(6,2)$ which gives me $90$ ways. But the textbook states $60$ ways. Can someone suggest how ?

Thanks.

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  • $\begingroup$ In each you have, altogether, $5\cdot 4\cdot 3$ possibilities, or $60$. $\endgroup$ – jm324354 Jun 1 '15 at 15:42
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    $\begingroup$ I think that your textbook is wrong. I also get $90$ numbers. $\endgroup$ – Dylan Jun 1 '15 at 15:44
  • $\begingroup$ $60$ is correct if you only have numbers $(1,2,3,4,5,6)$ to chose from. Or numbers $(2,3,4,5,6,7)$, $\endgroup$ – Joffan Jun 1 '15 at 23:46
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For the first digit you only have $3$ possibilities, $2,4$ or $6$. For the second digit, having chosen the first digit you only have $6$ possibilities left. For the third digit, having chosen the first two digit you have $5$ possibilities left. In total you have $3 \times 6 \times 5$ possibilities.

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    $\begingroup$ Normally you would call the rightmost digit 'last' or 'third', not 'first'. $\endgroup$ – user26486 Jun 1 '15 at 15:49
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    $\begingroup$ That depends on how you look at things. :) $\endgroup$ – Beni Bogosel Jun 1 '15 at 15:51
  • $\begingroup$ @BeniBogosel yeah right .thanks $\endgroup$ – Taylor Ted Jun 1 '15 at 15:51

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