Combinations of consecutive digits 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?
Thanks,
Kevin
 A: 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.
A: 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?
A: 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.;)
A: 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: $3!$
$$5! \cdot 3!$$
