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Find the number of passwords that use 3, 4, 5, 6, 7, 8, 9 exactly once.

I think I solved this part: it's 7!

Next question is: in how many of those 7! are the three even digits consecutive?

I been thinking about this for about an hour now and just can't get anywhere. Any advice?



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4 Answers

up vote 1 down vote accepted

There are just $3$ even digits. so you don't need to worry which of them form the consecutive group. Among all possible $7!$ passwords, these three can occupy any of the $\binom73=35$ triplets of positions, all of them being equally likely. Among those triplets, how many are consecutive? And what does this mean for the probability that a password has its even digits on a consecutive triplet of positions?

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Think of 4,6,8 as a single entity. That leaves you with 5 objects to be rearranged in 5! ways. Of course, 4,6,8 can also be rearrange in 3! ways.

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There are three evens($4,6,8$).

Take evens as one set, call it $k$

Now, you have to arrange $(k,3,5,7,9)$ . There has to be only arrangement which is consecutive(Obvious).

Therefore, the total arrangements will $5!1!$= $120$. Alas.;)

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Glue the three even numbers together. then you have essentially, 5 spots to fill with 5 objects:

$5!$ ways to do this.

Then there are $3!$ ways that $4, 6, 8$ can be ordered when glued together: $3!$

$$5! \cdot 3!$$

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Consecutive broda! . :) There are no arrangements within the glued ones. –  user63477 Mar 12 '13 at 5:54
The order in which those digits are glued can be done in 3! ways: they can be glued thusly: 468, 684, 846, 486, 648, 864. –  amWhy Mar 12 '13 at 5:56
Question is they are consecutive, so the only possible way is 468! –  user63477 Mar 12 '13 at 5:57
Consecutive means successively within their appearance in the password, not that the numbers must appear numbered consecutively. –  amWhy Mar 12 '13 at 5:59
He's asking in how many ways the three even digits are consecutive. So, it has to be $5!$, RIGHT? –  user63477 Mar 12 '13 at 6:05
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